【LeetCode】53. Maximum Subarray (2 solutions)

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.

 

解法一：

如果当前子串和小于等于0，则归零重新开始累加。记录最大子串和。

注意：ret需要初始化为INT_MIN（以防所有都为负）。

因此需要定义为long long类型。以防INT_MIN加上一个负数溢出。

class Solution {
public:
    int maxSubArray(int A[], int n) {
        long long ret = INT_MIN;
        long long cur = INT_MIN;
        for(int i = 0; i < n; i ++)
        {
            if(cur + A[i] > A[i])
                cur += A[i];
            else
                cur = A[i];
            ret = max(ret, cur);
        }
        return ret;
    }
};

 

或者初始化为第一个值

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int ret = nums[0];
        int cur = nums[0];
        for(int i = 1; i < nums.size(); i ++)
        {
            cur = max(nums[i], cur+nums[i]);
            ret = max(ret, cur);
        }
        return ret;
    }
};

 

解法二：分治法。将数组分成两段A1,A2之后，就分解成为子问题了：

1、最大子串在A1中；

2、最大子串在A2中；

3、最大子串是A1后缀+A2前缀。

class Solution {
public:
    int maxSubArray(int A[], int n) {
        if(n == 1)
            return A[0];
        else
        {
            int mid = n/2;
            //divide into [0~mid-1], [mid~n-1]
            int ret1 = maxSubArray(A, mid);
            int ret2 = maxSubArray(A+mid, n-mid);
            
            int left = mid-1;
            int ret3_1 = A[mid-1];
            int temp = A[mid-1];
            left --;
            while(left >= 0)
            {
                temp += A[left];
                ret3_1 = max(ret3_1, temp);
                left --;
            }
            
            int right = mid;
            int ret3_2 = A[mid];
            temp = A[mid];
            right ++;
            while(right < n)
            {
                temp += A[right];
                ret3_2 = max(ret3_2, temp);
                right ++;
            }
            
            return max(max(ret1, ret2), ret3_1+ret3_2);
        }
    }
};