Merge pull request #99 from KongJHong/master

yanglbme · web-flow · commit 2850ecaefd59 · 2018-11-12T18:34:28.000+08:00
Add Solution 123[CPP]
diff --git a/solution/0012.Integer to Roman/Solution.java b/solution/0012.Integer to Roman/Solution.java
@@ -6,4 +6,5 @@ class Solution {
     public String intToRoman(int num) {
         return M[num / 1000] + C[(num % 1000) / 100] + X[(num % 100) / 10] + I[num % 10];
     }
-}
+}
+
diff --git a/solution/0018.4Sum/Solution.cpp b/solution/0018.4Sum/Solution.cpp
@@ -0,0 +1,40 @@
+class Solution {
+public:
+    vector<vector<int>> fourSum(vector<int>& nums, int target) {
+        int len = nums.size();
+        vector<vector<int>> ans;
+        if(len < 4)return ans;
+        sort(nums.begin(),nums.end());
+        
+        int left;
+        int right;
+        int sum;
+        for(int i = 0;i<len;i++){
+            if(i> 0 && nums[i] == nums[i-1])continue;
+            for(int j = i + 1;j<len;j++){
+                if(j > i+1 && nums[j] == nums[j-1])continue;
+                
+                left = j+1;
+                right = len - 1;
+                
+                while(left < right){
+                    
+                    sum = nums[i] + nums[j] + nums[left] + nums[right];
+                    
+                    if(sum == target){
+                        vector<int> tmp{nums[i],nums[j],nums[left],nums[right]};
+                        ans.push_back(tmp);
+                        
+                        while(left < right && nums[left] == nums[left+1])left++;
+                        while(left < right && nums[right] == nums[right-1])right--;
+                        left++;
+                        right--;
+                    }
+                    else if(sum < target)left++;
+                    else if(sum > target)right--;
+                }       
+            }
+        }
+        return ans;
+    }
+};
diff --git a/solution/0031.Next Permutation/Solution.cpp b/solution/0031.Next Permutation/Solution.cpp
@@ -0,0 +1,20 @@
+class Solution {
+public:
+    void nextPermutation(vector<int> &nums) {
+	    int len = nums.size();
+
+	    if (len == 0)return;
+        int i,j;
+        for(i = len - 2;i>=0;i--){
+            if(nums[i+1] > nums[i]){
+                for(j = len - 1;j>i;j--){
+                    if(nums[j]>nums[i])break;
+                }   
+                swap(nums[i],nums[j]);
+                reverse(nums.begin()+i+1,nums.end());           
+                return;
+            }
+        }
+	    reverse(nums.begin(),nums.end());
+    }
+};
diff --git a/solution/0042.Trapping Rain Water/README.md b/solution/0042.Trapping Rain Water/README.md
@@ -0,0 +1,62 @@
+## 接雨水
+### 问题描述
+
+给定 n 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。
+
+![ ](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/22/rainwatertrap.png)
+
+上面是由数组`[0,1,0,2,1,0,1,3,2,1,2,1]` 表示的高度图，在这种情况下，可以接 6 个单位的雨水（蓝色部分表示雨水）。 感谢 Marcos 贡献此图。
+
+示例:
+```
+输入: [0,1,0,2,1,0,1,3,2,1,2,1]
+输出: 6
+```
+
+### 思路
+
+方法是找凹槽，怎么找呢？
+
+1. 设置`slow,fast`两个下标代表凹槽的左右边界，一旦遇到`height[fast]>=height[slow]`的情况，计算凹槽的容积
+2. 上面情况是以右边界高度一定大于左边界为准的，当形成凹槽且左边界大于右边界时，要怎么记录呢？答案是设置`stopPoint`点，规则是当`height[fast]>height[stopPoint]`时有`stopPoint = fast`记录右边最高点；同时当fast越界时，会到`stopPoint`上
+
+```CPP
+class Solution {
+public:
+    int trap(vector<int>& height) {
+        int len = height.size();
+        if(len == 0 || len == 1)return 0;
+        
+        int slow = 0;
+        int fast = 1;
+        
+        int stopPoint;
+        
+        int total = 0;
+        int bottom;
+        int tmp = 0;
+        while(fast < len){
+            //每次更新stopPoint
+            stopPoint = fast;
+            while(fast < len && height[fast] <= height[slow]){
+                if(height[fast] > height[stopPoint])
+                    stopPoint = fast;
+                fast++;
+            }
+            
+            //越界了要回到stopPoint
+            if(fast >= len)fast = stopPoint;
+
+            tmp = 0;
+            bottom = min(height[slow],height[fast]);
+            for(int i = slow+1;i<fast;i++){
+                tmp += bottom - height[i];
+            }
+            slow = fast;
+            total += tmp;
+            fast++;
+        }
+        return total;    
+    }  
+};
+```
diff --git a/solution/0042.Trapping Rain Water/Solution.cpp b/solution/0042.Trapping Rain Water/Solution.cpp
@@ -0,0 +1,38 @@
+class Solution {
+public:
+    int trap(vector<int>& height) {
+        int len = height.size();
+        if(len == 0 || len == 1)return 0;
+        
+        int slow = 0;
+        int fast = 1;
+        
+        int stopPoint;
+        
+        int total = 0;
+        int bottom;
+        int tmp = 0;
+        while(fast < len){
+            //每次更新stopPoint
+            stopPoint = fast;
+            while(fast < len && height[fast] <= height[slow]){
+                if(height[fast] > height[stopPoint])
+                    stopPoint = fast;
+                fast++;
+            }
+            
+            //越界了要回到stopPoint
+            if(fast >= len)fast = stopPoint;
+
+            tmp = 0;
+            bottom = min(height[slow],height[fast]);
+            for(int i = slow+1;i<fast;i++){
+                tmp += bottom - height[i];
+            }
+            slow = fast;
+            total += tmp;
+            fast++;
+        }
+        return total;    
+    }  
+};
diff --git a/solution/0123.Best Time to Buy and Sell Stock III/README.md b/solution/0123.Best Time to Buy and Sell Stock III/README.md
@@ -0,0 +1,86 @@
+## 买卖股票的最佳时机 III
+
+### 问题描述
+给定一个数组，它的第 i 个元素是一支给定的股票在第 i 天的价格。
+
+设计一个算法来计算你所能获取的最大利润。你最多可以完成 两笔 交易。
+
+注意: 你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。
+
+示例 1:
+```
+输入: [3,3,5,0,0,3,1,4]
+输出: 6
+解释: 在第 4 天（股票价格 = 0）的时候买入，在第 6 天（股票价格 = 3）的时候卖出，这笔交易所能获得利润 = 3-0 = 3 。
+     随后，在第 7 天（股票价格 = 1）的时候买入，在第 8 天 （股票价格 = 4）的时候卖出，这笔交易所能获得利润 = 4-1 = 3 。
+```
+示例 2:
+```
+输入: [1,2,3,4,5]
+输出: 4
+解释: 在第 1 天（股票价格 = 1）的时候买入，在第 5 天 （股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。   
+     注意你不能在第 1 天和第 2 天接连购买股票，之后再将它们卖出。   
+     因为这样属于同时参与了多笔交易，你必须在再次购买前出售掉之前的股票。
+```
+示例 3:
+```
+输入: [7,6,4,3,1] 
+输出: 0 
+解释: 在这个情况下, 没有交易完成, 所以最大利润为 0。
+```
+-------------
+### 思路：
+
+建立两个数组，因为是只能两次交易，所以有左右数组
+
+- 一个数组是`left[i]`,表示在第i天及之前交易的最大利润
+- 一个数组是`right[i]`,表示在第i天及之后交易的最大利润
+
+最后同时遍历，求和取出最大值就可以了
+
+```CPP
+class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        int left = 0;
+        int len = prices.size();
+        if(len == 0 || len == 1)return 0;
+        
+        vector<int> leftArr(len, 0);
+	    vector<int> rightArr(len, 0);
+
+	    int diff, day = 1, minPrice, maxPrice, maxProfit;
+
+	    //计算某一天及之前的最大利益
+	    minPrice = prices[0];
+	    maxProfit = 0;
+	    for (day = 1; day < len; day++) {
+		    diff = prices[day] - minPrice;
+		    if (diff < 0)minPrice = prices[day];
+		    else if (diff > maxProfit)maxProfit = diff;
+
+		    leftArr[day] = maxProfit;
+	    }
+
+	    //计算某一天及之前的最大利益
+	    maxPrice = prices[len - 1];
+	    maxProfit = 0;
+	    for (day = len - 2; day >= 0; day--) {
+		    diff = maxPrice - prices[day];
+		    if (diff < 0)maxPrice = prices[day];
+		    else if (diff > maxProfit)maxProfit = diff;
+
+		    rightArr[day] = maxProfit;
+        }
+
+	    int sum = 0;
+	    maxProfit = leftArr[0] + rightArr[0];
+	    for (int i = 1; i < len; i++) {
+		    sum = leftArr[i] + rightArr[i];
+		    if (sum > maxProfit)maxProfit = sum;
+	    }
+
+	    return maxProfit;
+    }
+};
+```
diff --git a/solution/0123.Best Time to Buy and Sell Stock III/Solution.cpp b/solution/0123.Best Time to Buy and Sell Stock III/Solution.cpp
@@ -0,0 +1,44 @@
+class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        int left = 0;
+        int len = prices.size();
+        if(len == 0 || len == 1)return 0;
+        
+        vector<int> leftArr(len, 0);
+	    vector<int> rightArr(len, 0);
+
+	    int diff, day = 1, minPrice, maxPrice, maxProfit;
+
+	    //计算某一天及之前的最大利益
+	    minPrice = prices[0];
+	    maxProfit = 0;
+	    for (day = 1; day < len; day++) {
+		    diff = prices[day] - minPrice;
+		    if (diff < 0)minPrice = prices[day];
+		    else if (diff > maxProfit)maxProfit = diff;
+
+		    leftArr[day] = maxProfit;
+	    }
+
+	    //计算某一天及之前的最大利益
+	    maxPrice = prices[len - 1];
+	    maxProfit = 0;
+	    for (day = len - 2; day >= 0; day--) {
+		    diff = maxPrice - prices[day];
+		    if (diff < 0)maxPrice = prices[day];
+		    else if (diff > maxProfit)maxProfit = diff;
+
+		    rightArr[day] = maxProfit;
+        }
+
+	    int sum = 0;
+	    maxProfit = leftArr[0] + rightArr[0];
+	    for (int i = 1; i < len; i++) {
+		    sum = leftArr[i] + rightArr[i];
+		    if (sum > maxProfit)maxProfit = sum;
+	    }
+
+	    return maxProfit;
+    }
+};
diff --git a/solution/0448.Find All Numbers Disappeared in an Array/README.md b/solution/0448.Find All Numbers Disappeared in an Array/README.md
@@ -0,0 +1,51 @@
+## 找到所有数组中消失的数字
+
+### 问题描述
+
+给定一个范围在  `1 ≤ a[i] ≤ n ( n = 数组大小 )` 的 整型数组，数组中的元素一些出现了两次，另一些只出现一次。
+
+找到所有在 [1, n] 范围之间没有出现在数组中的数字。
+
+您能在不使用额外空间且时间复杂度为O(n)的情况下完成这个任务吗? 你可以假定返回的数组不算在额外空间内。
+
+**示例:**
+```
+输入:
+[4,3,2,7,8,2,3,1]
+
+输出:
+[5,6]
+```
+
+### 解法
+
+题目要求不使用额外空间，所以计数排序的方法不可取；
+
+根据题目的条件，给定的`a[i]`是在`[1,n]`的范围内，所以可以利用这种关系来对数组进行处理，如`a[a[i]] = -a[a[i]`作反标记，最终若`a[i]>0`的话，则证明该下标`i`没有出现过，加入输出数组
+
+```CPP
+class Solution {
+public:
+    vector<int> findDisappearedNumbers(vector<int>& nums) {
+        int len = nums.size();
+        vector<int> ans;
+        if(len == 0)return ans;
+        
+        int index;
+        for(int i = 0;i<len;++i){
+            index = abs(nums[i]) - 1;
+            
+            if(nums[index] > 0)
+                nums[index] = -nums[index];   
+        }
+
+        
+        for(int i = 0;i<len;++i){
+            if(nums[i] > 0)
+                ans.push_back(i+1);
+        }
+        
+        return ans;
+    }
+};
+```
diff --git a/solution/0448.Find All Numbers Disappeared in an Array/Solution.cpp b/solution/0448.Find All Numbers Disappeared in an Array/Solution.cpp
@@ -0,0 +1,24 @@
+class Solution {
+public:
+    vector<int> findDisappearedNumbers(vector<int>& nums) {
+        int len = nums.size();
+        vector<int> ans;
+        if(len == 0)return ans;
+        
+        int index;
+        for(int i = 0;i<len;++i){
+            index = abs(nums[i]) - 1;
+            
+            if(nums[index] > 0)
+                nums[index] = -nums[index];   
+        }
+
+        
+        for(int i = 0;i<len;++i){
+            if(nums[i] > 0)
+                ans.push_back(i+1);
+        }
+        
+        return ans;
+    }
+};

Original file line number	Diff line number	Diff line change
`@@ -6,4 +6,5 @@ class Solution {`
`6`	`6`	`public String intToRoman(int num) {`
`7`	`7`	`return M[num / 1000] + C[(num % 1000) / 100] + X[(num % 100) / 10] + I[num % 10];`
`8`	`8`	`}`
`9`		`-}`
	`9`	`+}`
	`10`	`+`