[LeetCode] 410. Split Array Largest Sum

题目链接: https://leetcode.com/problems/split-array-largest-sum/description/

Description

Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays.

Note:
If n is the length of array, assume the following constraints are satisfied:

• 1 ≤ n ≤ 1000
• 1 ≤ m ≤ min(50, n)

Examples:

Input:
nums = [7,2,5,10,8]
m = 2

Output:
18

Explanation:
There are four ways to split nums into two subarrays.
The best way is to split it into [7,2,5] and [10,8],
where the largest sum among the two subarrays is only 18.

解题思路

动态规划，用 f[i][j] 表示前 i 个数分成 j 组的子组最大和的最小值。对于把 i 个数分成 j 组，先将前 k (k = 0, 1, …, i - 1) 个数划分 j - 1 组，最小的最大和为 f[k][j - 1]，余下的 [k + 1, i] 组成第 j 组计算和，两者之间较大值为当前分组的最大和，取这 k 种分组中最大和的最小值赋给 f[i][j]。计算从左上到右下，因此在计算 f[i][j] 时，f[k][j - 1] 已经计算好了，可以直接使用。

最终解为 f[m][n]。

Code

class Solution {
public:
    int splitArray(vector<int>& nums, int m) {
        int n = nums.size();

        vector<vector<int>> f(n + 1, vector<int>(m + 1, INT_MAX));
        vector<int> sub(n + 1, 0);

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

        f[0][0] = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                for (int k = 0; k < i; k++) {
                    f[i][j] = min(f[i][j], max(f[k][j - 1], sub[i] - sub[k]));
                }
            }
        }

        return f[n][m];
    }
};