leetcode53

题目：

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.

解法一：
这道题让我们求最大子数组之和，并且要我们用两种方法来解，分别是O(n)的解法，还有用分治法Divide and Conquer Approach，这个解法的时间复杂度是O(nlgn)，那我们就先来看O(n)的解法，定义两个变量res和curSum，其中res保存最终要返回的结果，即最大的子数组之和，curSum初始值为0，每遍历一个数字num，比较curSum + num和num中的较大值存入curSum，然后再把res和curSum中的较大值存入res，以此类推直到遍历完整个数组，可得到最大子数组的值存在res中。

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int res = INT_MIN, curSum = 0;
        for (int num : nums) {
            curSum = max(curSum + num, num);
            res = max(res, curSum);
        }
        return res;
    }
};

解法二：
题目还要求我们用分治法Divide and Conquer Approach来解，这个分治法的思想就类似于二分搜索法，我们需要把数组一分为二，分别找出左边和右边的最大子数组之和，然后还要从中间开始向左右分别扫描，求出的最大值分别和左右两边得出的最大值相比较取最大的那一个。

class Solution {
public:
    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));
    }
};

标签：nums,int,res,leetcode53,num,curSum,left
来源： https://blog.csdn.net/hua111hua/article/details/87926224