Leetcode: Maximum Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

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More practice:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

经典的连续子数组最大和，不过这里要求不能为空子数组。O(n)方法：

class Solution {public:    int maxSubArray(int A[], int n) {        int sum = 0;        int largest = numeric_limits<int>::min();        for (int i = 0; i < n; ++i) {            sum += A[i];            if (sum > largest) {                largest = sum;            }            if (sum < 0) {                sum = 0;            }        }                return largest;    }};

分治方法：

class Solution {public:    int maxSubArray(int A[], int n) {               return maxSubArrayUtil(A, 0 , n - 1);    }        int maxSubArrayUtil(int A[], int low, int up) {        if (low == up) {            return A[low];        }        else {            int mid = (low + up) >> 1;            int max_sum = max(maxSubArrayUtil(A, low, mid), maxSubArrayUtil(A, mid+1, up));                        int sum = 0;            int max_low = numeric_limits<int>::min();            for (int i = mid; i >= low; --i) {                sum += A[i];                if (sum > max_low) {                    max_low = sum;                }            }            sum = 0;            int max_up = numeric_limits<int>::min();            for (int i = mid + 1; i <= up; ++i) {                sum += A[i];                if (sum > max_up) {                    max_up = sum;                }            }                        return max(max_sum, max_low + max_up);        }    }};

按理说，分治方法时间复杂度为O(nlgn)，但测试用例时间花费反而小一些。可能跟测试集合有关系吧。