[LintCode] Maximum Subarray 最大子数组

 

Given an array of integers, find a contiguous subarray which has the largest sum.

 Notice

The subarray should contain at least one number.

Have you met this question in a real interview? 

Yes

Example

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

Challenge 

Can you do it in time complexity O(n)?

 

LeetCode上的原题，请参见我之前的博客Maximum Subarray。

 

解法一：

class Solution {
public:    
    /**
     * @param nums: A list of integers
     * @return: A integer indicate the sum of max subarray
     */
    int maxSubArray(vector<int> nums) {
        int res = INT_MIN, curSum = 0;
        for (int num : nums) {
            curSum += num;
            curSum = max(curSum, num);
            res = max(res, curSum);
        }
        return res;
    }
};

 

解法二：

class Solution {
public:    
    /**
     * @param nums: A list of integers
     * @return: A integer indicate the sum of max subarray
     */
    int maxSubArray(vector<int> nums) {
        if (nums.empty()) return 0;
        return helper(nums, 0, (int)nums.size() - 1);
    }
    int helper(vector<int>& nums, int left, int right) {
        if (left >= right) return nums[left];
        int mid = left + (right - left) / 2;
        int lmax = helper(nums, left, mid - 1);
        int rmax = helper(nums, mid + 1, right);
        int mmax = nums[mid], t = mmax;
        for (int i = mid - 1; i >= left; --i) {
            t += nums[i];
            mmax = max(mmax, t);
        }
        t = mmax;
        for (int i = mid + 1; i <= right; ++i) {
            t += nums[i];
            mmax = max(mmax, t);
        }
        return max(mmax, max(lmax, rmax));
    }
};