a) Maximum subarray sum in left half b) Maximum subarray sum in right half c) Maximum subarray sum such that the subarray crosses the midpoint*/returnMath.Max(Math.Max(maxSubArraySum(arr, l, m), maxSubArraySum(arr, m+1, h)), maxCrossingSum(arr, l, m, h)); }/*Driver program t...
class Solution { public int maxSubArray(int[] nums) { int pre = 0, maxAns = nums[0]; for (int x : nums) { pre = Math.max(pre + x, x); maxAns = Math.max(maxAns, pre); } return maxAns; } } 方法2:分治 class Solution { public class Status { public int lSum, rSum, ...
the contiguous subarray[4,−1,2,1]has the largest sum =6. 分析:这道题和152题有点类似。是求连续的子数组的和的最大值。是一道动态规划的题目。 代码如下: publicclassSolution {publicintMaxSubArray(int[] nums) {intmax=int.MinValue,sum=int.MinValue;for(inti=0;i<nums.Length;i++) {if(...
题目描述:点击此处经典的字段和问题 1 class Solution { 2 public: 3 int maxSubArray(int A[], int n) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int m
## LeetCode 53 最大子数列和 Maximum Subarray class Solution(): def maxSubArray(self, nums): l = len(nums) dp = [0] * l ## 初始化数组全部为0 ## 套路第三步,初始特殊值为 nums 第一个元素 dp[0] = nums[0] #res_max = dp[0] ## 最终结果也初始化为 nums 第一个元素 for i in...
classSolution{public:intmaxSubArray(vector<int>& nums){intn = nums.size();if(n ==0)return0;intret = nums[0], sum =0, min_sum =0;for(inti =0; i < n; ++i) { sum += nums[i]; ret =max(ret, sum - min_sum); min_sum =min(min_sum, sum); ...
Solution { public: int maxSubArray(vector<int>& nums) { int res = INT_MIN; int tempSum = 0; for(auto i:nums){ tempSum += i; if(tempSum > res) res = tempSum; if(tempSum < 0) tempSum = 0; } return res; } };_
int k = solution.maxSubArray(nums); // 主功能 // 输出 cout << k << endl; } 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 3.2 功能程序 3.2.1 穷举遍历法 (1)代码 #pragma once #include<vector>// std::vector ...
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle. 翻译: 给你一个整数型数组nums,找出连续的子数组(长度至少为1)使得其和最大,返回这个最大值。 举例: 输入:[-2,1,-3,4,-1,2,1,-5,4] ...
classSolution {public:intmaxSubArray(vector<int>&nums) {intlen=nums.size();intmax_ending_here=nums[0];intmax_so_far=nums[0];for(inti=1; i<len; i++) { max_ending_here=max(nums[i],max_ending_here+nums[i]); max_so_far=max(max_so_far,max_ending_here); ...