[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 = ;
        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 ;
        return helper(nums, , (int)nums.size() - );
    }
    int helper(vector<int>& nums, int left, int right) {
        if (left >= right) return nums[left];
        int mid = left + (right - left) / ;
        int lmax = helper(nums, left, mid - );
        int rmax = helper(nums, mid + , right);
        int mmax = nums[mid], t = mmax;
        for (int i = mid - ; i >= left; --i) {
            t += nums[i];
            mmax = max(mmax, t);
        }
        t = mmax;
        for (int i = mid + ; i <= right; ++i) {
            t += nums[i];
            mmax = max(mmax, t);
        }
        return max(mmax, max(lmax, rmax));
    }
};