[LeetCode] 410. Split Array Largest Sum

題目鏈接: 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.

解題思路

動態規劃，用 f[i][j] 表示前 i 個數分成 j 組的子組最大和的最小值。對於把 i 個數分成 j 組，先將前 k (k = 0, 1, …, i - 1) 個數劃分 j - 1 組，最小的最大和爲 f[k][j - 1]，餘下的 [k + 1, i] 組成第 j 組計算和，兩者之間較大值爲當前分組的最大和，取這 k 種分組中最大和的最小值賦給 f[i][j]。計算從左上到右下，因此在計算 f[i][j] 時，f[k][j - 1] 已經計算好了，可以直接使用。

最終解爲 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];
    }
};