leetcode 410. Split Array Largest Sum

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本文介绍了一种算法，用于将一个由非负整数组成的数组分成m个连续的非空子数组，并使这些子数组的最大和最小化。通过动态规划的方法，详细解释了如何有效地解决这个问题。

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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.

class Solution {
public:
    long splitArray(vector<int>& nums, int m) 
    {
        int n = nums.size();
        sum = vector<long>(n,0);
        sum[0] = nums[0];
        mem = vector<vector<long>>(n,vector<long>(m+1,INT_MAX));
        for(int i=1;i<n;i++)
        {
            sum[i] = sum[i-1] + nums[i];
        }
        return splitarray(nums,n,m);
    }
private:
    vector<long> sum;
    vector<vector<long>> mem;
    long  splitarray(vector<int>& num,int n ,int m)
    {
        for(int i=0;i<n;i++)
            for(int j=1;j<=i+1&&j<=m;j++)
                if(j==1) mem[i][j] = sum[i];
                else
                    for(int k=0;k<i;k++) mem[i][j] = min(mem[i][j],max(mem[k][j-1],sum[i]-sum[k])); 
        return mem[n-1][m];
    }
};