[Lintcode]Maximum Subarray III最大子数组 III

Given an array of integers and a number k, find k 

non-overlapping

 subarrays
which have the largest sum.

The number in each subarray should be contiguous.

Return the largest sum.
Example

Given 

[-1,4,-2,3,-2,3]

, 

k=2

,
return 

8

分析：与Best Time to Buy and Sell Stock IV类似，两个数组分别记录包含当前值的本地最优解和全局最优解。local[i][j]代表从0到nums中第i个数字处，分成j个数组后，和的最大值,且必须包括第i个数字，这样可以排除第i个数组包含在前一个数组中的情况。global[i][j]代表从0到nums第i个数字处，分成j个数组后和的最大值。
二维数组动态规划的状态转移方程为：

local[i][j] = Math.max(local[i - 1][j], global[i - 1][j - 1])  + nums[i - 1];

global[i][j] = Math.max(global[i - 1][j], local[i][j]);

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) {
if(k > nums.length) return 0;

int[][] local = new int[nums.length + 1][k + 1];
int[][] global = new int[nums.length + 1][k + 1];
//if k=0 local/global colomn 0 = 0
//if nums.length=0 local/global row 0 = 0
for(int i = 1; i <= nums.length; i++) {
local[i][0] = Integer.MIN_VALUE;
for(int j = 1; j <= k; j++) {
if(j > i) {//矩阵中的值不能默认为0，否则影响结果
local[i][j] = Integer.MIN_VALUE;
global[i][j] = Integer.MIN_VALUE;
continue;
}
local[i][j] = Math.max(local[i - 1][j], global[i - 1][j - 1]) + nums[i - 1];
if(i == j)
global[i][j] = local[i][j];
else
global[i][j] = Math.max(global[i - 1][j], local[i][j]);
}
}

return global[nums.length][k];
}
}