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题目

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.

Example 1:

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.

我的想法

用双指针，哪边的和小就移动哪边的指针。但是如果输入为负数呢？？？

class Solution {
   	//记录一下错误做法    public int pivotIndex(int[] nums) {
           if(nums == null || nums.length <= 1) return -1;        int l = 0, r = nums.length-1;        int sumL = nums[l], sumR = nums[r];        while(r - l > 2){
               if(sumL > sumR){
                   r--;                sumR += nums[r];            }             else{
                   l++;                sumL += nums[l];            }         }        if(sumL == sumR) return l+1;        else return -1;    }}

解答

只用一个指针，将sum分成两半，如果左右两半的和不相同则移动指针

class Solution {
       public int pivotIndex(int[] nums) {
           if(nums == null || nums.length <= 1) return -1;        int leftsum = 0, sum = 0;        for(int i = 0; i

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