[Leetcode] Maximum Subarray

描述

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.

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.

找出数组中和最大的子数组，返回最大值。

分析1

很容易想到动态规划的方法解决这道题。我们令 dpi 表示以 ai 为结尾的子数组中的最大和，那么我们要求的就是所有 dp 中的最大值。

递推公式是：dpi+1=max(dpi+ai+1,ai+1)

代码1

class Solution {public:    int maxSubArray(vector<int>& nums) {        if (nums.empty()) return INT_MIN;        vector<int> dp(nums.size(), 0);        dp[0] = nums[0];        int res = dp[0];        for (int i = 1; i < nums.size(); i++) {            dp[i] = max(nums[i], nums[i] + dp[i - 1]);            if (res < dp[i]) res = dp[i];        }        return res;    }};

分析2

还有一种分而治之的思路，数组 [lo,hi] 的子数组，无非以下三种情况之一：
1. 属于左半部分 [lo,mid−1]
2. 属于左半部分 [mid+1,hi]
3. 包含 [mid]

对于前两种情况，可以直接递归求解，对于第三种情况，可以将子数组分成三部分 [lo,mid−1]，[mid]，[mid+1,high] ，因此可以以 mid 为中心向左右扩展查找最大和，代码如下。

代码2

class Solution {public:    int maxSubArray(vector<int>& nums) {        int res = INT_MIN;        return maxArray(nums, 0, nums.size() - 1, res);    }    int maxArray(vector<int>& nums, int lo, int hi, int& res){        if (lo > hi) return INT_MIN;        int mid = lo + (hi - lo) / 2;        int lmax = maxArray(nums, lo, mid - 1, res);        int rmax = maxArray(nums, mid + 1, hi, res);        res = max(max(lmax, rmax), res);        int sum = 0, mlmax = 0, mrmax = 0;        for (int i = mid - 1; i >= lo; i--) {            sum += nums[i];            if (sum > mlmax) mlmax = sum;        }        sum = 0;        for (int i = mid + 1; i <= hi; i++) {            sum += nums[i];            if (sum > mrmax) mrmax = sum;        }        res = max(res, mlmax + mrmax + nums[mid]);        return res;    }};