Lintcode - Maximum Subarray III

Given an array of integers and a number k, find knon-overlapping subarrays which have the largest sum.

The number in each subarray should be contiguous.

Return the largest sum.

Note

The subarray should contain at least one number

Example

Given [-1,4,-2,3,-2,3],k=2, return 8

d[i][j]代表0->i-1元素中j个subarray的maxsum  (注意不包含元素i)

d[i][j] = max(d[i][j], d[m][j-1] + max) (m = j-1 .... i-1; max需要单独求，是从元素i-1到m的max subarray, 用求max subarray的方法，需要从后往前算）

    public int maxSubArray(ArrayList nums, int k) {
        // write your code
        int n = nums.size();
        int[][] d = new int[n+1][k+1];
        for (int j = 1; j <= k; j++) {
            for (int i = j; i <= n; i++) {
                d[i][j] = Integer.MIN_VALUE;
                int max = Integer.MIN_VALUE;
                int localMax = 0;
                for (int m = i-1; m >= j-1; m--) {
                    localMax = Math.max(nums.get(m), nums.get(m)+localMax);
                    max = Math.max(localMax, max);
                    d[i][j] = Math.max(d[i][j], d[m][j-1] + max);
                }
            }
        }
        return d[n][k];
    }