***[Lintcode]Maximum Subarray Difference 最大子數組差

Given an array with integers.

Find two non-overlapping subarrays A and B, which |SUM(A) - SUM(B)| is the largest.

Return the largest difference.

Example

For [1, 2, -3, 1], return 6.

分析：求四次最大/最小子字符串。分別爲從左到右計算最大/最小連續子字符串，從右至左計算最大/最小連續子字符串。原理類似於動態規劃，如

leftMin[i] = Math.min(leftMin[i - 1], Math.min(nums[i] + min, nums[i])); leftMin代表0到i處（包括）連續最小子字符串和。其等於index=i-1時leftMin的值，或者包含或者不包含nums[i]的sum。

最後遍歷一次計算leftMin和rightMax 或者leftMax和rightMin的絕對值的最大值。注意錯位求和。

public class Solution {
    /**
     * @param nums: A list of integers
     * @return: An integer indicate the value of maximum difference between two
     *          Subarrays
     */
    public int maxDiffSubArrays(int[] nums) {
        int[] leftMin = new int[nums.length];
        int[] leftMax = new int[nums.length];
        //i向前連續n項元素的和
        int min = 0, max = 0;
        for(int i = 0; i < nums.length; i++) {
            min = Math.min(nums[i] + min, nums[i]);
            max = Math.max(nums[i] + max, nums[i]);
            if(i == 0) {
                leftMin[i] = min;
                leftMax[i] = max;
            } else {
                leftMin[i] = Math.min(leftMin[i - 1], min);
                leftMax[i] = Math.max(leftMax[i - 1], max);
            }
        }
        int res = 0, endMin = 0, endMax = 0;
        min = 0;max = 0;
        for(int i = nums.length - 1; i > 0; i--) {
            min = Math.min(nums[i] + min, nums[i]);
            max = Math.max(nums[i] + max, nums[i]);
            if(i == nums.length - 1) {
                endMin = min;
                endMax = max;
            } else {
                endMin = Math.min(endMin, min);
                endMax = Math.max(endMax, max);
            }
            res = Math.max(Math.max(Math.abs(leftMax[i-1]-endMin),   
                Math.abs(endMax - leftMin[i-1])), res);  
        }
        return res;
    }

}