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Aug 5, 2023
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feat: add solutions to lcof2 problem: No.012 #1398

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118 changes: 74 additions & 44 deletions lcof2/剑指 Offer II 012. 左右两边子数组的和相等/README.md
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Expand Up @@ -60,7 +60,20 @@

<!-- 这里可写通用的实现逻辑 -->

用前缀和进行预处理，避免重复计算
**方法一：前缀和**

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 $left = 0$, $right = \sum_{i = 0}^{n - 1} nums[i]$。

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。

遍历结束，如果没有找到中间位置，返回 $-1$。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 $nums$ 的长度。

相似题目：

- [1991. 找到数组的中间位置](/solution/1900-1999/1991.Find%20the%20Middle%20Index%20in%20Array/README.md)
- [2574. 左右元素和的差值](/solution/2500-2599/2574.Left%20and%20Right%20Sum%20Differences/README.md)

<!-- tabs:start -->

Expand All @@ -71,13 +84,12 @@
```python
class Solution:
def pivotIndex(self, nums: List[int]) -> int:
n = len(nums)
sum = [0 for _ in range(n + 1)]
for i in range(1, n + 1):
sum[i] = sum[i - 1] + nums[i - 1]
for i in range(0, n):
if sum[i] == sum[n] - sum[i + 1]:
left, right = 0, sum(nums)
for i, x in enumerate(nums):
right -= x
if left == right:
return i
left += x
return -1
```

Expand All @@ -88,59 +100,79 @@ class Solution:
```java
class Solution {
public int pivotIndex(int[] nums) {
int n = nums.length;
int[] sum = new int[n + 1];
for (int i = 1; i <= n; i++) {
sum[i] = sum[i - 1] + nums[i - 1];
int left = 0, right = 0;
for (int x : nums) {
right += x;
}
for (int i = 0; i < n; i++) {
if (sum[i] == sum[n] - sum[i + 1]) {
int n = nums.length;
for (int i = 0; i < n; ++i) {
right -= nums[i];
if (left == right) {
return i;
}
left += nums[i];
}
return -1;
}
}
```

### **C++**

```cpp
class Solution {
public:
int pivotIndex(vector<int>& nums) {
int left = 0;
int right = accumulate(nums.begin(), nums.end(), 0);
int n = nums.size();
for (int i = 0; i < n; ++i) {
right -= nums[i];
if (left == right) {
return i;
}
left += nums[i];
}
return -1;
}
};
```

### **Go**

```go
func pivotIndex(nums []int) int {
n := len(nums)
sum := make([]int, n+1)
for i := 1; i <= n; i++ {
sum[i] = sum[i-1] + nums[i-1]
left, right := 0, 0
for _, x := range nums {
right += x
}
for i := 0; i < n; i++ {
if sum[i] == sum[n]-sum[i+1] {
for i, x := range nums {
right -= x
if left == right {
return i
}
left += x
}
return -1
}
```

### **C++**

```cpp
class Solution {
public:
int pivotIndex(vector<int>& nums) {
int sum = 0;
int total = 0;
for (int num : nums)
sum += num;

for (int i = 0; i < nums.size(); i++) {
total += nums[i];
if (total - nums[i] == sum - total)
return i;
### **TypeScript**

```ts
function pivotIndex(nums: number[]): number {
let left = 0;
let right = nums.reduce((a, b) => a + b, 0);
const n = nums.length;
for (let i = 0; i < n; ++i) {
right -= nums[i];
if (left === right) {
return i;
}

return -1;
left += nums[i];
}
};
return -1;
}
```

### **PHP**
Expand All @@ -152,16 +184,14 @@ class Solution {
* @return Integer
*/
function pivotIndex($nums) {
$sum = 0;
$pre = 0;
for ($i = 0; $i < count($nums); $i++) {
$sum += $nums[$i];
}
$left = 0;
$right = array_sum($nums);
for ($i = 0; $i < count($nums); $i++) {
if ($pre === $sum - $pre - $nums[$i]) {
$right -= $nums[$i];
if ($left == $right) {
return $i;
}
$pre += $nums[$i];
$left += $nums[$i];
}
return -1;
}
Expand Down
17 changes: 8 additions & 9 deletions lcof2/剑指 Offer II 012. 左右两边子数组的和相等/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,17 +1,16 @@
class Solution {
public:
int pivotIndex(vector<int>& nums) {
int sum = 0;
int total = 0;
for (int num : nums)
sum += num;

for (int i = 0; i < nums.size(); i++) {
total += nums[i];
if (total - nums[i] == sum - total)
int left = 0;
int right = accumulate(nums.begin(), nums.end(), 0);
int n = nums.size();
for (int i = 0; i < n; ++i) {
right -= nums[i];
if (left == right) {
return i;
}
left += nums[i];
}

return -1;
}
};
13 changes: 7 additions & 6 deletions lcof2/剑指 Offer II 012. 左右两边子数组的和相等/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,13 +1,14 @@
func pivotIndex(nums []int) int {
n := len(nums)
sum := make([]int, n+1)
for i := 1; i <= n; i++ {
sum[i] = sum[i-1] + nums[i-1]
left, right := 0, 0
for _, x := range nums {
right += x
}
for i := 0; i < n; i++ {
if sum[i] == sum[n]-sum[i+1] {
for i, x := range nums {
right -= x
if left == right {
return i
}
left += x
}
return -1
}
32 changes: 17 additions & 15 deletions lcof2/剑指 Offer II 012. 左右两边子数组的和相等/Solution.java
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Original file line number Diff line number Diff line change
@@ -1,15 +1,17 @@
class Solution {
public int pivotIndex(int[] nums) {
int n = nums.length;
int[] sum = new int[n + 1];
for (int i = 1; i <= n; i++) {
sum[i] = sum[i - 1] + nums[i - 1];
}
for (int i = 0; i < n; i++) {
if (sum[i] == sum[n] - sum[i + 1]) {
return i;
}
}
return -1;
}
}
class Solution {
public int pivotIndex(int[] nums) {
int left = 0, right = 0;
for (int x : nums) {
right += x;
}
int n = nums.length;
for (int i = 0; i < n; ++i) {
right -= nums[i];
if (left == right) {
return i;
}
left += nums[i];
}
return -1;
}
}
12 changes: 5 additions & 7 deletions 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 number Diff line number Diff line change
Expand Up @@ -4,16 +4,14 @@ class Solution {
* @return Integer
*/
function pivotIndex($nums) {
$sum = 0;
$pre = 0;
$left = 0;
$right = array_sum($nums);
for ($i = 0; $i < count($nums); $i++) {
$sum += $nums[$i];
}
for ($i = 0; $i < count($nums); $i++) {
if ($pre === $sum - $pre - $nums[$i]) {
$right -= $nums[$i];
if ($left == $right) {
return $i;
}
$pre += $nums[$i];
$left += $nums[$i];
}
return -1;
}
Expand Down
19 changes: 9 additions & 10 deletions lcof2/剑指 Offer II 012. 左右两边子数组的和相等/Solution.py
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Original file line number Diff line number Diff line change
@@ -1,10 +1,9 @@
class Solution:
def pivotIndex(self, nums: List[int]) -> int:
n = len(nums)
sum = [0 for _ in range(n + 1)]
for i in range(1, n + 1):
sum[i] = sum[i - 1] + nums[i - 1]
for i in range(0, n):
if sum[i] == sum[n] - sum[i + 1]:
return i
return -1
class Solution:
def pivotIndex(self, nums: List[int]) -> int:
left, right = 0, sum(nums)
for i, x in enumerate(nums):
right -= x
if left == right:
return i
left += x
return -1
13 changes: 13 additions & 0 deletions 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 number Diff line number Diff line change
@@ -0,0 +1,13 @@
function pivotIndex(nums: number[]): number {
let left = 0;
let right = nums.reduce((a, b) => a + b, 0);
const n = nums.length;
for (let i = 0; i < n; ++i) {
right -= nums[i];
if (left === right) {
return i;
}
left += nums[i];
}
return -1;
}