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feat: add solutions to lc problem: No.1642
No.1642.Furthest Building You Can Reach
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‎solution/1600-1699/1642.Furthest Building You Can Reach/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心 + 优先队列（小根堆）**
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梯子最好用在高度差较大的地方，因此我们可以将所有的高度差存入优先队列中，每次取出最小的高度差，如果梯子不够用，则用砖块填补，如果砖块不够用，则返回当前位置。
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时间复杂度 $O(n\log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 `heights` 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
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h = []
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for i, a in enumerate(heights[:-1]):
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b = heights[i + 1]
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d = b - a
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if d > 0:
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heappush(h, d)
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if len(h) > ladders:
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bricks -= heappop(h)
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if bricks < 0:
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return i
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return len(heights) - 1
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int furthestBuilding(int[] heights, int bricks, int ladders) {
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PriorityQueue<Integer> q = new PriorityQueue<>();
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int n = heights.length;
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for (int i = 0; i < n - 1; ++i) {
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int a = heights[i], b = heights[i + 1];
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int d = b - a;
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if (d > 0) {
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q.offer(d);
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if (q.size() > ladders) {
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bricks -= q.poll();
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if (bricks < 0) {
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return i;
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}
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}
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}
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}
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return n - 1;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int furthestBuilding(vector<int>& heights, int bricks, int ladders) {
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priority_queue<int, vector<int>, greater<int>> q;
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int n = heights.size();
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for (int i = 0; i < n - 1; ++i) {
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int a = heights[i], b = heights[i + 1];
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int d = b - a;
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if (d > 0) {
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q.push(d);
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if (q.size() > ladders) {
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bricks -= q.top();
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q.pop();
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if (bricks < 0) {
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return i;
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}
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}
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}
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}
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return n - 1;
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}
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};
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```
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### **Go**
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```go
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func furthestBuilding(heights []int, bricks int, ladders int) int {
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q := hp{}
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n := len(heights)
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for i, a := range heights[:n-1] {
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b := heights[i+1]
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d := b - a
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if d > 0 {
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heap.Push(&q, d)
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if q.Len() > ladders {
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bricks -= heap.Pop(&q).(int)
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if bricks < 0 {
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return i
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}
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}
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}
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}
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return n - 1
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}
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type hp struct{ sort.IntSlice }
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func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) }
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func (h *hp) Pop() interface{} {
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a := h.IntSlice
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v := a[len(a)-1]
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h.IntSlice = a[:len(a)-1]
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return v
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}
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```
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### **...**

‎solution/1600-1699/1642.Furthest Building You Can Reach/README_EN.md

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@@ -62,13 +62,104 @@ It is impossible to go beyond building 4 because you do not have any more bricks
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### **Python3**
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```python
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class Solution:
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def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
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h = []
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for i, a in enumerate(heights[:-1]):
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b = heights[i + 1]
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d = b - a
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if d > 0:
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heappush(h, d)
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if len(h) > ladders:
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bricks -= heappop(h)
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if bricks < 0:
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return i
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return len(heights) - 1
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```
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### **Java**
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```java
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class Solution {
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public int furthestBuilding(int[] heights, int bricks, int ladders) {
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PriorityQueue<Integer> q = new PriorityQueue<>();
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int n = heights.length;
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for (int i = 0; i < n - 1; ++i) {
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int a = heights[i], b = heights[i + 1];
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int d = b - a;
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if (d > 0) {
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q.offer(d);
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if (q.size() > ladders) {
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bricks -= q.poll();
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if (bricks < 0) {
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return i;
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}
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}
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}
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}
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return n - 1;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int furthestBuilding(vector<int>& heights, int bricks, int ladders) {
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priority_queue<int, vector<int>, greater<int>> q;
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int n = heights.size();
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for (int i = 0; i < n - 1; ++i) {
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int a = heights[i], b = heights[i + 1];
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int d = b - a;
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if (d > 0) {
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q.push(d);
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if (q.size() > ladders) {
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bricks -= q.top();
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q.pop();
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if (bricks < 0) {
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return i;
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}
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}
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}
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}
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return n - 1;
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}
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};
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```
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### **Go**
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```go
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func furthestBuilding(heights []int, bricks int, ladders int) int {
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q := hp{}
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n := len(heights)
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for i, a := range heights[:n-1] {
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b := heights[i+1]
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d := b - a
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if d > 0 {
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heap.Push(&q, d)
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if q.Len() > ladders {
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bricks -= heap.Pop(&q).(int)
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if bricks < 0 {
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return i
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}
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}
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}
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}
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return n - 1
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}
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type hp struct{ sort.IntSlice }
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func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) }
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func (h *hp) Pop() interface{} {
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a := h.IntSlice
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v := a[len(a)-1]
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h.IntSlice = a[:len(a)-1]
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return v
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}
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```
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### **...**

‎solution/1600-1699/1642.Furthest Building You Can Reach/Solution.cpp

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class Solution {
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public:
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int furthestBuilding(vector<int>& heights, int bricks, int ladders) {
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priority_queue<int, vector<int>, greater<int>> q;
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int n = heights.size();
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for (int i = 0; i < n - 1; ++i) {
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int a = heights[i], b = heights[i + 1];
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int d = b - a;
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if (d > 0) {
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q.push(d);
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if (q.size() > ladders) {
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bricks -= q.top();
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q.pop();
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if (bricks < 0) {
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return i;
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}
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}
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}
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}
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return n - 1;
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}
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};

‎solution/1600-1699/1642.Furthest Building You Can Reach/Solution.go

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func furthestBuilding(heights []int, bricks int, ladders int) int {
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q := hp{}
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n := len(heights)
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for i, a := range heights[:n-1] {
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b := heights[i+1]
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d := b - a
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if d > 0 {
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heap.Push(&q, d)
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if q.Len() > ladders {
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bricks -= heap.Pop(&q).(int)
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if bricks < 0 {
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return i
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}
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}
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}
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}
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return n - 1
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}
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type hp struct{ sort.IntSlice }
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func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) }
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func (h *hp) Pop() interface{} {
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a := h.IntSlice
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v := a[len(a)-1]
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h.IntSlice = a[:len(a)-1]
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return v
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}

‎solution/1600-1699/1642.Furthest Building You Can Reach/Solution.java

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class Solution {
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public int furthestBuilding(int[] heights, int bricks, int ladders) {
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PriorityQueue<Integer> q = new PriorityQueue<>();
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int n = heights.length;
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for (int i = 0; i < n - 1; ++i) {
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int a = heights[i], b = heights[i + 1];
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int d = b - a;
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if (d > 0) {
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q.offer(d);
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if (q.size() > ladders) {
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bricks -= q.poll();
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if (bricks < 0) {
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return i;
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}
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}
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}
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}
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return n - 1;
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}
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}

‎solution/1600-1699/1642.Furthest Building You Can Reach/Solution.py

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class Solution:
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def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
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h = []
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for i, a in enumerate(heights[:-1]):
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b = heights[i + 1]
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d = b - a
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if d > 0:
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heappush(h, d)
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if len(h) > ladders:
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bricks -= heappop(h)
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if bricks < 0:
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return i
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return len(heights) - 1

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