LWC 58：724. Find Pivot Index

LWC 58：724. Find Pivot Index

传送门：724. Find Pivot Index

Problem:

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.

Example 2:

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].

思路：
判断当前坐标点的左侧元素之和是否和右侧元素之和相等，返回第一个符合情况的下标。

空间换时间，说白了，以O(1)的速度查询累加和，以O(n)的速度更新累加和数组即可，再以O(n)的速度搜索符合情况的下标。

代码如下：

    public int pivotIndex(int[] nums) {
        int n = nums.length;
        int[] lf = new int[n];
        int[] rt = new int[n];

        int sum = 0;
        for (int i = 0; i < n; ++i) {
             lf[i] = sum;
             sum += nums[i];
        }

        sum = 0;
        for (int j = n - 1; j >= 0; --j) {
            rt[j] = sum;
            sum += nums[j];
        }

        for (int i = 0; i < n; ++i) {
            if (lf[i] == rt[i]) {
                return i;
            }
        }

        return -1;
    }

累加和版本：

    public int pivotIndex(int[] nums) {
        int n = nums.length;
        int[] sums = new int[n + 1];

        for (int i = 0; i < n; ++i){
            sums[i + 1] = sums[i] + nums[i];
        }

        for (int i = 0; i < n; ++i){
            if (sums[i + 1] == sums[n] - sums[i]){
                return i;
            }
        }

        return -1;
    }