【LeetCode】548. Split Array with Equal Sum 解题报告(C++)

• 作者： 负雪明烛
• id： fuxuemingzhu
• 个人博客：http://fuxuemingzhu.cn/

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目录

• • 题目描述
  • 题目大意
  • 解题方法
  • • 暴力
  • 日期

题目地址：https://leetcode-cn.com/problems/split-array-with-equal-sum/

题目描述

Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies following conditions:

1. 0 < i, i + 1 < j, j + 1 < k < n - 1
2. Sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) should be equal.

where we define that subarray (L, R) represents a slice of the original array starting from the element indexed L to the element indexed R.

Example:

Input: [1,2,1,2,1,2,1]
Output: True
Explanation:
i = 1, j = 3, k = 5.
sum(0, i - 1) = sum(0, 0) = 1
sum(i + 1, j - 1) = sum(2, 2) = 1
sum(j + 1, k - 1) = sum(4, 4) = 1
sum(k + 1, n - 1) = sum(6, 6) = 1

Note:

• 1 <= n <= 2000.
• Elements in the given array will be in range [-1,000,000, 1,000,000].

题目大意

在nums数组中插入三个隔板i,j,k，使得 (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1)的和相等。

解题方法

暴力

这个题好像没有简单的做法，直接暴力三层循环即可。一个常见的优化就是提前算出从位置0到每个位置的累加和preSum，这样区间[i, j]的和 = preSum[j] - preSum[i - 1];

另外有个case是1000多个0，导致超时，此时需要一个优化：跳过nums[j] == nums[j-1] == 0的点。

C++代码如下：

class Solution {
public:
    bool splitArray(vector<int>& nums) {
        const int N = nums.size();
        vector<long long> preSum(N, 0);
        long long sum = 0;
        for (int i = 0; i < N; ++i) {
            sum += nums[i];
            preSum[i] = sum;
        }
        for (int i = 1; i < N; ++i) {
            long long sum_0i = preSum[i - 1];
            for (int j = i + 1; j < N; ++j) {
                long long sum_ij = preSum[j - 1] - preSum[i];
                if ((nums[j] == 0 && nums[j - 1] == 0) || sum_0i != sum_ij)
                    continue;
                for (int k = j + 1; k < N; ++k) {
                    long long sum_jk = preSum[k - 1] - preSum[j];
                    if (sum_0i != sum_jk)
                        continue;
                    long long sum_k = preSum[N - 1] - preSum[k];
                    if (sum_0i == sum_k) {
                        return true;
                    }
                }
            }
        }
        return false;
    }
};

日期

2019 年 9 月 20 日 —— 是选择中国互联网式加班？还是外企式养生？