寻找数组的中心索引

题目

class Solution {public int pivotIndex(int[] nums) {for (int i = 0; i< nums.length; i++){int left = 0;        //每次进行重置int right = 0;for (int j = i+1; j<nums.length; j++) {right += nums[j];}for (int j = i-1; j>=0; j--){left += nums[j];}if (left == right) {return i;    //返回索引；如果有多个符合值，直接返回第一个，不需要把所有符合值都找出来                   } }return -1;      //不用加if，如果有中心下标，已经返回i}
}

class Solution {public int pivotIndex(int[] nums) {int totalnum = 0;for (int num : nums) {totalnum += num;}for (int i = 0; i< nums.length; i++){int left = 0;int right = 0;for (int j = i+1; j<nums.length; j++) {right += nums[j];}left = totalnum - right - nums[i];if (left == right) {return i;} }return -1;}
}

class Solution {public int pivotIndex(int[] nums) {int totalnum = 0;for (int num : nums) {totalnum += num;}int left = 0;int right = 0;for (int i = 0; i< nums.length; i++){left = totalnum - right - nums[i];if (left == right) {return i;} right += nums[i];}return -1;}
}

class Solution {public int pivotIndex(int[] nums) {int totalnum = 0;for (int num : nums) {totalnum += num;}int right = 0;for (int i = 0; i< nums.length; i++){if (2*right + nums[i] == totalnum) {    //left == rightreturn i;} right += nums[i];}return -1;}
}

总结：

效率优化：先计算总和，再动态更新两侧和

明确返回：数组下标指的是索引

代码冗余：循环结束，代表没有找到数组下标，直接返回-1，不需要再加条件

判断条件：left == right,  2*left + nums[i] = totalnum

提示条件：根据提示条件可得，数组范围【10^-7, 10^7】, int类型数据范围【（-2）^31, 2^31 - 1】,所以使用int类型数组即可。