[Leetcode] Split Array Largest Sum

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.

这道题目，可以采用递归，那么思路如下：

dfs(int m,int i){    if(边界条件) ...    int min = Integer.MAX_VALUE:    for(int j = i + 1; j < len; j++){        max = Math.max(sum(i,j), dfs(m+1, j));        min = Math.min(min,max);    }    return min;}

m表示当前切了几次，i表示从哪个元素开始切。改成dp以后，可以采用dp[m][i] 表示从i开始一直到结尾的数组分成m组每组和最大值最小。
从而有一个递推，将i以后切成m段，这m段实机包含一个sum(i,j) 和dp[m-1][j] ，也就是一段加上 m-1段。找出所有情况的最小值。
dp[m][i] = min { max{ sum[i][j], dp[m-1][j]}。
实现的时候，一个要注意每一片都不能为空，所以i和j的最大值有界限。同时可以对数组进行预处理，获得一个累加和数组sum，从而 nums[i] + nums[i+1]…+nums[j] = sum[j] - sum[i]。

public class Solution {    public int splitArray(int[] nums, int m) {        int[][] dp = new int[m][nums.length];        int len = nums.length;        int[] sum = new int[len + 1];        sum[0] = 0;        for (int i = 0; i < len; i++) {            sum[i + 1] = sum[i] + nums[i];        }        for(int i=0; i<len; i++)            dp[0][i] = sum[len]-sum[i];        for (int k = 1; k < m; k++) {            for (int i = 0;i < len - k;i ++) {                dp[k][i] =  Integer.MAX_VALUE;                for (int j = i + 1; j <= len - k;j++) {                    dp[k][i] = Math.min(dp[k][i], Math.max(sum[j] - sum[i], dp[k - 1][j]));                }            }        }        return dp[m - 1][0];    }}

dp的空间可以进一步优化。因为我们改变的只是一个dp[k][i]的值，而使用的是它上一行i后面的元素，所以可以直接优化进一个一维数组中。代码如下：

public class Solution {    public int splitArray(int[] nums, int m) {        int[] dp = new int[nums.length];        int len = nums.length;        int[] sum = new int[len + 1];        sum[0] = 0;        for (int i = 0; i < len; i++) {            sum[i + 1] = sum[i] + nums[i];        }        for(int i=0; i<len; i++)            dp[i] = sum[len]-sum[i];        for (int k = 1; k < m; k++) {            for (int i = 0;i < len - k;i ++) {                dp[i] =  Integer.MAX_VALUE;                for (int j = i + 1; j <= len - k;j++) {                    dp[i] = Math.min(dp[i], Math.max(sum[j] - sum[i], dp[j]));                }            }        }        return dp[0];    }}

这道题目，采用递归，化为循环dp，优化空间三个步骤。