19.2.8 [LeetCode 53] Maximum Subarray

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example:

Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.

Follow up:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

题意

求连续子集的最大和

题解

其实我觉得这题挺难的|||

首先是O(n)解法

 1 class Solution {
 2 public:
 3     int maxSubArray(vector<int>& nums) {
 4         int e = 1, n = nums.size(), ans = nums[0], now = nums[0];
 5         while (e < n) {
 6             now = max(now + nums[e], nums[e]);
 7             ans = max(now, ans);
 8             e++;
 9         }
10         return ans;
11     }
12 };

然后是分治算法

 1 class Solution {
 2 public:
 3     int maxSubRange(vector<int>&nums, int x, int y) {
 4         if (x == y)return nums[x];
 5         int mid = (x + y) / 2;
 6         int lmax = maxSubRange(nums, x, mid), rmax = maxSubRange(nums, mid + 1, y);
 7         int tmp = 0, mmax = 0;
 8         for (int i = mid - 1; i >= x; i--) {
 9             tmp += nums[i];
10             mmax = max(tmp, mmax);
11         }
12         tmp = mmax;
13         for (int i = mid; i <= y; i++) {
14             tmp += nums[i];
15             mmax = max(tmp, mmax);
16         }
17         return max(mmax, max(lmax, rmax));
18     }
19     int maxSubArray(vector<int>& nums) {
20         int e = 1, n = nums.size(), ans = nums[0], now = nums[0];
21         while (e < n) {
22             now = max(now + nums[e], nums[e]);
23             ans = max(now, ans);
24             e++;
25         }
26         return ans;
27     }
28 };

原文地址：https://www.cnblogs.com/yalphait/p/10356399.html