标签：src code while turn style solution coding another ide
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.
求连续子集的最大和
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 };
19.2.8 [LeetCode 53] Maximum Subarray
标签：src code while turn style solution coding another ide
原文地址：https://www.cnblogs.com/yalphait/p/10356399.html
