# C#LeetCode刷题之#724-寻找数组的中心索引（ Find Pivot Index）

nums 的长度范围为 [0, 10000]。

Given an array of integers nums, write a method that returns the “pivot” index of this array.

We define the pivot index as the index where the sum of the numbers to the left of the index is equal to the sum of the numbers to the right of the index.

If no such index exists, we should return -1. If there are multiple pivot indexes, you should return the left-most pivot index.

Input: nums = [1, 7, 3, 6, 5, 6]

Output: 3

Explanation: The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3.
Also, 3 is the first index where this occurs.

Input: nums = [1, 2, 3]

Output: -1

Explanation: There is no index that satisfies the conditions in the problem statement.

Note:

The length of nums will be in the range [0, 10000].
Each element nums[i] will be an integer in the range [-1000, 1000].

```public class Program {

public static void Main(string[] args) {
int[] nums = null;

nums = new int[] { 1, 7, 3, 6, 5, 6 };
var res = PivotIndex(nums);
Console.WriteLine(res);

nums = new int[] { 1, 2, 3 };
res = PivotIndex2(nums);
Console.WriteLine(res);

nums = new int[] { -1, -1, -1, 0, 0, 1, 1 };
res = PivotIndex3(nums);
Console.WriteLine(res);

}

private static int PivotIndex(int[] nums) {
//暴力解法
for(int i = 0; i < nums.Length; i++) {
if(ComputerLeft(nums, i) == ComputerRight(nums, i)) {
return i;
}
}
return -1;
}

private static int ComputerLeft(int[] nums, int index) {
int sum = 0;
for(int i = 0; i < index; i++) {
sum += nums[i];
}
return sum;
}

private static int ComputerRight(int[] nums, int index) {
int sum = 0;
for(int i = index + 1; i < nums.Length; i++) {
sum += nums[i];
}
return sum;
}

private static int PivotIndex2(int[] nums) {
//和数组解法，先存下所有之前的数的和再分析
if(nums.Length == 0) return -1;
int[] sums = new int[nums.Length];
sums[0] = nums[0];
for(int i = 1; i < nums.Length; i++) {
sums[i] = sums[i - 1] + nums[i];
}
//边界处理
if(sums[nums.Length - 1] == sums[0]) return 0;
for(int i = 1; i < nums.Length; i++) {
//比较之前的和、之前的和是否相等
if(sums[i - 1] == sums[nums.Length - 1] - sums[i]) {
return i;
}
}
return -1;
}

private static int PivotIndex3(int[] nums) {
//解法和思路同下面注释掉的代码部分
int sum = 0;
for(int k = 0; k < nums.Length; k++) {
sum += nums[k];
}
int leftSum = 0;
int i = 0, j = nums.Length - 1;
while(i <= j) {
if((sum -= nums[i]) == leftSum) return i;
leftSum += nums[i++];
}
return -1;
//int leftSum = 0;
//for(int i = 0; i < nums.Length; i++) {
//    sum -= nums[i];
//    if(leftSum == sum)
//        return i;
//    leftSum += nums[i];
//}
//return -1;
}

}```

```3
-1
0```