Lintcode ：45. 最大子数组差

描述

给定一个整数数组，找出两个不重叠的子数组A和B，使两个子数组和的差的绝对值|SUM(A) - SUM(B)|最大。

返回这个最大的差值。

子数组最少包含一个数

样例

给出数组[1, 2, -3, 1]，返回 6

挑战

时间复杂度为O(n)，空间复杂度为O(n)

思路：与上一题最大子数组II类似，只是要分情况找到sum(A)最大，sum(B)最小和sum(A)最小，sum(B)最大的情况

public int maxDiffSubArrays(int[] nums) {
        // write your code here
        int[] leftMax = new int[nums.length];
        int[] leftMin = new int[nums.length];
        int[] rightMin = new int[nums.length];
        int[] rightMax = new int[nums.length];
        int curSum = 0;
        int maxSum = nums[0];

        int curDiff = 0;
        int minSum = nums[0];
        for (int i = 0; i < nums.length; i++) {
            curSum += nums[i];
            maxSum = Math.max(maxSum, curSum);
            curSum = Math.max(curSum, 0);
            leftMax[i] = maxSum;

            curDiff += nums[i];
            minSum = Math.min(minSum, curDiff);
            curDiff = Math.min(curDiff, 0);
            leftMin[i] = minSum;
        }

        curSum = 0;
        curDiff = 0;
        maxSum = nums[nums.length - 1];
        minSum = nums[nums.length - 1];
        for (int i = nums.length - 1; i >= 0; i--) {
            curDiff += nums[i];
            minSum = Math.min(minSum, curDiff);
            curDiff = Math.min(curDiff, 0);
            rightMin[i] = minSum;

            curSum += nums[i];
            maxSum = Math.max(maxSum, curSum);
            curSum = Math.max(curSum, 0);
            rightMax[i] = maxSum;
        }

        int result = 0;
        for (int i = 0; i < nums.length - 1; i++) {
            result = Math.max(result, Math.abs(leftMax[i] - rightMin[i + 1]));
            result = Math.max(result, Math.abs(leftMin[i] - rightMax[i + 1]));
        }


        return result;
    }