Maximum Subarray--Leetcode

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

for example, given the Array [?2,1,?3,4,?1,2,1,?5,4" ,
the contiguous subarray < Code style= "Font-family:menlo,monaco,consolas, ' Courier New ', monospace; font-size:12.6000003814697px; PADDING:2PX 4px; Color:rgb (199,37,78); Background-color:rgb (249,242,244) ">[4,?1,2,1"  has the largest sum = 6 .

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More Practice:

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

Algorithm One

O (n), the actual execution time on the Leetcode is 16ms.

Class Solution {public:    int maxsubarray (int a[], int n) {        if (n <= 0) return int_min;        int sum = a[0];        int maxsum = sum;        for (int i=1; i<n; i++) {                sum = max (A[i], sum+a[i]);                Maxsum = max (maxsum, sum);        }        return maxsum;}    ;

Algorithm two divide and conquer

O (Nlogn), the actual execution time on the Leetcode is 18ms.

Class Solution {public:    int maxsubarray (int a[], int n) {        if (n <= 0) return int_min;        Return helper (A, 0, n-1);    }    int helper (int a[], int left, int.) {        if (left = right)                return a[left];        const int MID = left + (right-left)/2;        int sum = A[mid];        int midmax = sum;        for (int i=mid-1; i>=left; i--) {                sum + = A[i];                Midmax = max (Midmax, sum);        }        sum = Midmax;        for (int i=mid+1; i<=right; i++) {                sum + = A[i];                Midmax = max (Midmax, sum);        }        const int Leftmax = Helper (A, left, mid);        const int Rightmax = Helper (A, mid+1, right);        Return Max (Midmax, Max (Leftmax, Rightmax));}    ;

Reference from:

Https://oj.leetcode.com/discuss/694/how-solve-maximum-subarray-using-divide-and-conquer-approach

Maximum Subarray--Leetcode