[LintCode]Maximum Subarray III

http://www.lintcode.com/en/problem/maximum-subarray-iii/#

找出k个不重叠的子数组，且和最大

不太明显的DP，dp[i][j]为分成i个子数组，原数组长度为j。外层i遍历k，内层遍历从i ~ len。内层的local为当前分割之后的最后一个子数组的结尾位置为nums[j - 1]。

当i == j的时候注意一下

public class Solution {    /**     * @param nums: A list of integers     * @param k: An integer denote to find k non-overlapping subarrays     * @return: An integer denote the sum of max k non-overlapping subarrays     */    public int maxSubArray(int[] nums, int k) {        // write your code here        if (nums == null || nums.length < k) {            return 0;        }        int[][] dp = new int[k + 1][nums.length + 1];        for (int i = 1; i <= k; i++) {            int local = Integer.MIN_VALUE;            for (int j = i; j <= nums.length; j++) {                local = Math.max(local, dp[i - 1][j - 1]) + nums[j - 1];                if (j == i) {                    dp[i][j] = local;                } else {                    dp[i][j] = Math.max(local, dp[i][j - 1]);                }            }        }        return dp[k][nums.length];    }}