leetcode-1013-easy

Partition Array Into Three Parts With Equal Sum

Given an array of integers arr, return true if we can partition the array into three non-empty parts with equal sums.

Formally, we can partition the array if we can find indexes i + 1 < j with (arr[0] + arr[1] + ... + arr[i] == arr[i + 1] + arr[i + 2] + ... + arr[j - 1] == arr[j] + arr[j + 1] + ... + arr[arr.length - 1])

Example 1:

Input: arr = [0,2,1,-6,6,-7,9,1,2,0,1]
Output: true
Explanation: 0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1
Example 2:

Input: arr = [0,2,1,-6,6,7,9,-1,2,0,1]
Output: false
Example 3:

Input: arr = [3,3,6,5,-2,2,5,1,-9,4]
Output: true
Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
Constraints:

3 <= arr.length <= 5 * 104
-104 <= arr[i] <= 104

思路一：数组分三段，分别统计这三段的值，注意每段求和的时候需要提前赋值

    public static boolean canThreePartsEqualSum(int[] arr) {
        int sum = 0;
        for (int i : arr) {
            sum += i;
        }

        if (sum % 3 != 0) {
            return false;
        }

        int val = sum / 3;

        int begin = 0;
        int end = arr.length - 1;

        int partOne = arr[begin++];
        while (partOne != val && begin <= end) {
            partOne += arr[begin];
            begin++;
        }

        int partThree = arr[end--];
        while (partThree != val && end >= 0) {
            partThree += arr[end];
            end--;
        }

        if (begin > end) {
            return false;
        }

        int parTwo = arr[begin++];
        while (begin <= end) {
            parTwo += arr[begin];
            begin++;
        }

        return partOne == parTwo && parTwo == partThree;
    }

思路二：看了一下题解，中间那段不用计算，如果首尾的和相等，中间那段也一定相等

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