LeetCode Maximum Subarray

class Solution {
public:
    int maxSubArray(int A[], int n) {
        int cur_sum = 0;
        int max_sum = INT_MIN;
        
        for (int i=0; i<n; i++) {
            cur_sum += A[i];
            if (cur_sum > max_sum) max_sum = cur_sum;
            if (cur_sum < 0) {
                cur_sum = 0;
            }
        }
        return max_sum;
    }
};

老题

第二轮：

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.

click to show more practice.

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.

理解就不用记了，更简洁一些的做法依照公式：f(n) = max(f(n-1) + A[0], A[0])

class Solution {
public:
    int maxSubArray(int A[], int n) {
        if (A == NULL || n < 1) {
            return INT_MIN;
        }
        int contsum = A[0];
        int maxsum = A[0];
        
        for (int i=1; i<n; i++) {
            contsum = max(contsum + A[i], A[i]);
            maxsum = max(maxsum, contsum);
        }
        return maxsum;
    }
};

分治法，nlogn, 但是TLE

class Solution {
public:
int maxSubArray(int A[], int n) {
        return dfs(A, 0, n);
    }
    
    int dfs(int A[], int start, int end) {
        if (start >= end) {
            cout<<"range error: start="<<start<<", end="<<end<<endl;
            return INT_MIN;
        }
        if (start + 1 == end) return A[start];
        int mid = (start + end) / 2;
        int ma = dfs(A, start, mid);
        int mb = dfs(A, mid, end);
        int maxsum = ma > mb ? ma : mb;
        int lsum = 0, rsum = 0;
        int lmax = INT_MIN, rmax = INT_MIN;
        
        for (int i=mid - 1; i>=0; i--) {
            lsum += A[i];
            if (lsum > lmax) lmax = lsum;
        }
        
        for (int i=mid; i<end; i++) {
            rsum += A[i];
            if (rsum > rmax) rmax = rsum;
        }
        
        if (lmax + rmax > maxsum) maxsum = lmax + rmax;
        return maxsum;
    };
};