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feat: add solutions to lcof2 problem: No.012 (#1398)
NO.012.左右两边子数组的和相等
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‎lcof2/剑指 Offer II 012. 左右两边子数组的和相等/README.md

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Original file line numberDiff line numberDiff line change
@@ -60,7 +60,20 @@
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<!-- 这里可写通用的实现逻辑 -->
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用前缀和进行预处理，避免重复计算
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**方法一：前缀和**
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我们定义变量 $left$ 表示数组 $nums$ 中下标 $i$ 左侧元素之和，变量 $right$ 表示数组 $nums$ 中下标 $i$ 右侧元素之和。初始时 $left = 0$, $right = \sum_{i = 0}^{n - 1} nums[i]$。
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遍历数组 $nums$，对于当前遍历到的数字 $x$，我们更新 $right = right - x$，此时如果 $left=right$，说明当前下标 $i$ 就是中间位置，直接返回即可。否则，我们更新 $left = left + x$，继续遍历下一个数字。
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遍历结束，如果没有找到中间位置，返回 $-1$。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 $nums$ 的长度。
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相似题目：
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- [1991. 找到数组的中间位置](/solution/1900-1999/1991.Find%20the%20Middle%20Index%20in%20Array/README.md)
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- [2574. 左右元素和的差值](/solution/2500-2599/2574.Left%20and%20Right%20Sum%20Differences/README.md)
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<!-- tabs:start -->
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@@ -71,13 +84,12 @@
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```python
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class Solution:
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def pivotIndex(self, nums: List[int]) -> int:
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n = len(nums)
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sum = [0 for _ in range(n + 1)]
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for i in range(1, n + 1):
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sum[i] = sum[i - 1] + nums[i - 1]
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for i in range(0, n):
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if sum[i] == sum[n] - sum[i + 1]:
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left, right = 0, sum(nums)
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for i, x in enumerate(nums):
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right -= x
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if left == right:
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return i
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left += x
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return -1
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```
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@@ -88,59 +100,79 @@ class Solution:
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```java
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class Solution {
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public int pivotIndex(int[] nums) {
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int n = nums.length;
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int[] sum = new int[n + 1];
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for (int i = 1; i <= n; i++) {
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sum[i] = sum[i - 1] + nums[i - 1];
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int left = 0, right = 0;
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for (int x : nums) {
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right += x;
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}
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for (int i = 0; i < n; i++) {
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if (sum[i] == sum[n] - sum[i + 1]) {
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int n = nums.length;
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for (int i = 0; i < n; ++i) {
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right -= nums[i];
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if (left == right) {
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return i;
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}
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left += nums[i];
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}
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return -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 pivotIndex(vector<int>& nums) {
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int left = 0;
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int right = accumulate(nums.begin(), nums.end(), 0);
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int n = nums.size();
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for (int i = 0; i < n; ++i) {
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right -= nums[i];
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if (left == right) {
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return i;
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}
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left += nums[i];
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}
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return -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 pivotIndex(nums []int) int {
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n := len(nums)
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sum := make([]int, n+1)
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for i := 1; i <= n; i++ {
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sum[i] = sum[i-1] + nums[i-1]
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left, right := 0, 0
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for _, x := range nums {
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right += x
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}
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for i := 0; i < n; i++ {
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if sum[i] == sum[n]-sum[i+1] {
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for i, x := range nums {
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right -= x
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if left == right {
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return i
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}
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left += x
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}
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return -1
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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 pivotIndex(vector<int>& nums) {
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int sum = 0;
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int total = 0;
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for (int num : nums)
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sum += num;
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for (int i = 0; i < nums.size(); i++) {
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total += nums[i];
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if (total - nums[i] == sum - total)
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return i;
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### **TypeScript**
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```ts
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function pivotIndex(nums: number[]): number {
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let left = 0;
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let right = nums.reduce((a, b) => a + b, 0);
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const n = nums.length;
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for (let i = 0; i < n; ++i) {
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right -= nums[i];
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if (left === right) {
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return i;
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}
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return -1;
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left += nums[i];
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}
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};
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return -1;
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}
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```
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### **PHP**
@@ -152,16 +184,14 @@ class Solution {
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* @return Integer
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*/
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function pivotIndex($nums) {
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$sum = 0;
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$pre = 0;
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for ($i = 0; $i < count($nums); $i++) {
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$sum += $nums[$i];
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}
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$left = 0;
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$right = array_sum($nums);
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for ($i = 0; $i < count($nums); $i++) {
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if ($pre === $sum - $pre - $nums[$i]) {
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$right -= $nums[$i];
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if ($left == $right) {
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return $i;
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}
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$pre += $nums[$i];
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$left += $nums[$i];
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}
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return -1;
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}

‎lcof2/剑指 Offer II 012. 左右两边子数组的和相等/Solution.cpp

Lines changed: 8 additions & 9 deletions
Original file line numberDiff line numberDiff line change
@@ -1,17 +1,16 @@
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class Solution {
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public:
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int pivotIndex(vector<int>& nums) {
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int sum = 0;
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int total = 0;
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for (int num : nums)
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sum += num;
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for (int i = 0; i < nums.size(); i++) {
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total += nums[i];
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if (total - nums[i] == sum - total)
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int left = 0;
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int right = accumulate(nums.begin(), nums.end(), 0);
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int n = nums.size();
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for (int i = 0; i < n; ++i) {
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right -= nums[i];
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if (left == right) {
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return i;
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}
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left += nums[i];
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}
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return -1;
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}
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};

‎lcof2/剑指 Offer II 012. 左右两边子数组的和相等/Solution.go

Lines changed: 7 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -1,13 +1,14 @@
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func pivotIndex(nums []int) int {
2-
n := len(nums)
3-
sum := make([]int, n+1)
4-
for i := 1; i <= n; i++ {
5-
sum[i] = sum[i-1] + nums[i-1]
2+
left, right := 0, 0
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for _, x := range nums {
4+
right += x
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}
7-
for i := 0; i < n; i++ {
8-
if sum[i] == sum[n]-sum[i+1] {
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for i, x := range nums {
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right -= x
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if left == right {
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return i
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}
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left += x
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}
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return -1
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}

‎lcof2/剑指 Offer II 012. 左右两边子数组的和相等/Solution.java

Lines changed: 17 additions & 15 deletions
Original file line numberDiff line numberDiff line change
@@ -1,15 +1,17 @@
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class Solution {
2-
public int pivotIndex(int[] nums) {
3-
int n = nums.length;
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int[] sum = new int[n + 1];
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for (int i = 1; i <= n; i++) {
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sum[i] = sum[i - 1] + nums[i - 1];
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}
8-
for (int i = 0; i < n; i++) {
9-
if (sum[i] == sum[n] - sum[i + 1]) {
10-
return i;
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}
12-
}
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return -1;
14-
}
15-
}
1+
class Solution {
2+
public int pivotIndex(int[] nums) {
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int left = 0, right = 0;
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for (int x : nums) {
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right += x;
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}
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int n = nums.length;
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for (int i = 0; i < n; ++i) {
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right -= nums[i];
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if (left == right) {
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return i;
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}
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left += nums[i];
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}
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return -1;
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}
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}

‎lcof2/剑指 Offer II 012. 左右两边子数组的和相等/Solution.php

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Original file line numberDiff line numberDiff line change
@@ -4,16 +4,14 @@ class Solution {
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* @return Integer
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*/
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function pivotIndex($nums) {
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$sum = 0;
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$pre = 0;
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$left = 0;
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$right = array_sum($nums);
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for ($i = 0; $i < count($nums); $i++) {
10-
$sum += $nums[$i];
11-
}
12-
for ($i = 0; $i < count($nums); $i++) {
13-
if ($pre === $sum - $pre - $nums[$i]) {
10+
$right -= $nums[$i];
11+
if ($left == $right) {
1412
return $i;
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}
16-
$pre += $nums[$i];
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$left += $nums[$i];
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}
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return -1;
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}

‎lcof2/剑指 Offer II 012. 左右两边子数组的和相等/Solution.py

Lines changed: 9 additions & 10 deletions
Original file line numberDiff line numberDiff line change
@@ -1,10 +1,9 @@
1-
class Solution:
2-
def pivotIndex(self, nums: List[int]) -> int:
3-
n = len(nums)
4-
sum = [0 for _ in range(n + 1)]
5-
for i in range(1, n + 1):
6-
sum[i] = sum[i - 1] + nums[i - 1]
7-
for i in range(0, n):
8-
if sum[i] == sum[n] - sum[i + 1]:
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return i
10-
return -1
1+
class Solution:
2+
def pivotIndex(self, nums: List[int]) -> int:
3+
left, right = 0, sum(nums)
4+
for i, x in enumerate(nums):
5+
right -= x
6+
if left == right:
7+
return i
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left += x
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return -1

‎lcof2/剑指 Offer II 012. 左右两边子数组的和相等/Solution.ts

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,13 @@
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function pivotIndex(nums: number[]): number {
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let left = 0;
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let right = nums.reduce((a, b) => a + b, 0);
4+
const n = nums.length;
5+
for (let i = 0; i < n; ++i) {
6+
right -= nums[i];
7+
if (left === right) {
8+
return i;
9+
}
10+
left += nums[i];
11+
}
12+
return -1;
13+
}

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