[LintCode] Maximum Subarray Difference

Maximum Subarray Difference

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

Note

The subarray should contain at least one number

Challenge

O(n) time and O(n) space.

 

左右两头扫，注意要分别记录左边最大子段和、最小子段和，右边最大子段和、最小子段和。

class Solution {
public:
    /**
     * @param nums: A list of integers
     * @return: An integer indicate the value of maximum difference between two
     *          Subarrays
     */
    int maxDiffSubArrays(vector<int> nums) {
        // write your code here
        if (nums.empty()) return -1;
        int N = nums.size();
        vector<int> max_sum_l(N), min_sum_l(N);
        int max_tmp = nums[0], min_tmp = nums[0];
        max_sum_l[0] = min_sum_l[0] = nums[0];
        for (int i = 1; i < N; ++i) {
            max_tmp = max(max_tmp + nums[i], nums[i]);
            min_tmp = min(min_tmp + nums[i], nums[i]);
            max_sum_l[i] = max(max_tmp, max_sum_l[i-1]);
            min_sum_l[i] = min(min_tmp, min_sum_l[i-1]);
        }
        vector<int> max_sum_r(N), min_sum_r(N);
        max_tmp = min_tmp = nums[N-1];
        max_sum_r[N-1] = min_sum_r[N-1] = nums[N-1];
        for (int i = N - 2; i >= 0; --i) {
            max_tmp = max(max_tmp + nums[i], nums[i]);
            min_tmp = min(min_tmp + nums[i], nums[i]);
            max_sum_r[i] = max(max_tmp, max_sum_r[i+1]);
            min_sum_r[i] = min(min_tmp, min_sum_r[i+1]);
        }
        int res = 0;
        for (int i = 0; i < N - 1; ++i) {
            res = max(res, abs(max_sum_l[i] - min_sum_r[i+1]));
            res = max(res, abs(min_sum_l[i] - max_sum_r[i+1]));
        }
        return res;
    }
};