LeetCode刷题(C++)——Maximum Subarray(Easy)
来源:互联网 发布:java微信开发教程csdn 编辑:程序博客网 时间:2024/06/14 00:52
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
.思路:
(1)暴力枚举法:起点i=0...n-1,终点j=i....n-1,然后求和i...j,时间复杂度为O(n^3)
(2)枚举(优化):起点i=0...n-1,终点j=i...n-1,同时求和i...j,时间复杂度为(O(n^2)
(3)分治法:
分:将数组分成两个基本等长的子数组,分别求解T(n/2)
合:跨越中心点的最大子数组(枚举),时间复杂度为O(n)
总的时间复杂度为O(nlogn)
代码如下:
class Solution {public: int maxSubArray(vector<int>& nums) { int n = nums.size(); if (n==0) return 0; if (n == 1) return nums[0]; return maxSubArray(nums, 0, n - 1); } int maxSubArray(vector<int>& nums, int start, int end) { if (start > end) return INT_MIN; int n = end-start+1; if (n == 1) return nums[start]; int mid = (start+end)>>1; //分成两部分:0.....mid | mid+1......n-1 int answer = max(maxSubArray(nums,start,mid-1), maxSubArray(nums,mid+1,end)); int now = nums[mid], may = now; for (int i = mid - 1;i >= 0; i--) { may = max(may, now += nums[i]); } now = may; for (int i = mid+1;i <= end;i++) { may = max(may, now +=nums[i]); } return max(answer, may); }};
(4)利用动态规划思想
假设dp[i]表示以nums[i]结尾的最大子数组的和
dp[i]=max(dp[i-1]+nums[i],nums[i])
最大子数组包含nums[i-1]:dp[i-1]+nums[i]
最大子数组不包含nums[i-1]:nums[i]
两者当中取最大值
初值dp[0]=nums[0]
最大子数组的和为dp[0]...dp[n-1]中最大者
时间复杂度为O(n),空间复杂度为O(n)
代码如下:
class Solution {public: int maxSubArray(vector<int>& nums) { int n=nums.size(); vector<int> dp(n); dp[0]=nums[0]; int answer = dp[0]; for(int i=1; i<n;i++){ dp[i]=max(dp[i-1]+nums[i],nums[i]); answer=max(answer,dp[i]); } return answer; }};
同时还可以对dp进行空间优化:使得空间复杂度为O(1),用一个变量endHere来存储目前最大的子数组和
endHere = max(endHere+nums[i],nums[i]) // 等号左边的endHere相当于dp[i],等号右边的相当于dp[i-1]
answer = max(endHere,answer)
代码如下:
class Solution {public: int maxSubArray(vector<int>& nums) { int n=nums.size(); int endHere = nums[0]; int answer = nums[0]; for(int i=1; i<n;i++){ endHere=max(endHere+nums[i],nums[i]); answer=max(answer,endHere); } return answer; }};
(5)线性枚举
定义sum[i]=nums[0]+nums[1]+....+nums[i]; i>=0
则对于i...j的和:nums[i]+nums[i+1]+...+nums[j]=sum[j]-sum[i-1]
对于当前的sum[j],当sum[i-1]取最小值时,sum[j]才是最大值,于是我们可以用一个变量来记录sum的最小值
时间复杂度为O(n),空间复杂度为O(1)
代码如下:
class Solution {public: int maxSubArray(vector<int>& nums) { int n=nums.size(); int sum = nums[0]; int minSum = min(sum,0); int answer = nums[0]; for(int i=1; i<n;i++) { sum+=nums[i]; answer = max(answer, sum-minSum); minSum = min(sum,minSum); } return answer; }};
设置一个变量subSum记录nums[i]之前子数组和,最大子数组记为sum
当subSum<=0时,subSum+nums[i]<=nums[i],此时令subSum=nums[i],
当subSum>0时,subSum+nums[i]>nums[i]
比较:sum = max(sum,subSum)
最后返回sum
代码如下:
class Solution {public: int maxSubArray(vector<int>& nums) { int n=nums.size(); if(n==0) return 0; int subSum = 0; int sum = INT_MIN; for(int i=0; i<n;i++) { if(subSum<=0) subSum = nums[i]; else subSum+=nums[i]; if(sum<subSum) sum = subSum; } return sum; }};
1 0
- LeetCode刷题(C++)——Maximum Subarray(Easy)
- LeetCode 53. Maximum Subarray(Easy)
- LeetCode53. Maximum Subarray(easy)
- LeetCode-Easy刷题(11) Maximum Subarray
- 算法分析与设计丨第二周丨LeetCode(4)——Maximum Subarray(Easy)
- 【Leetcode-Easy-53】Maximum Subarray
- LeetCode 53. Maximum Subarray (Easy)
- LeetCode刷题(C++)——Maximum Depth of Binary Tree(Easy)
- 【leetcode】Array—— Maximum Subarray(53)
- 【leetcode】Array—— Maximum Product Subarray(152)
- Leetcode学习(22)—— Maximum Subarray
- LeetCode 53 — Maximum Subarray(C++ Java Python)
- LeetCode——Maximum Subarray
- LeetCode——Maximum Subarray
- leetcode——Maximum Subarray
- LeetCode——Maximum Subarray
- LeetCode——Maximum subarray
- LeetCode—Maximum Product Subarray
- 详解JUC之原子类概述
- Python自定义豆瓣电影种类,排行,点评的爬取与存储(基础)
- 使用STM32CubeMX创建STM32F407工程
- centos shell 编程-通过端口号kill对应的进程
- 鸡仔单片机成长记----------------学会使用宏晶ISP软件
- LeetCode刷题(C++)——Maximum Subarray(Easy)
- css3选择器
- 用Redis bitmap统计活跃用户、留存
- Arduino分割字符串
- Java多线程基础解析
- Linux命令行刻录光盘、光盘转ISO文件、校验光盘
- ContentProvider内容提供者 和 ContentResolver内容解析者
- fat文件系统
- 搭建完全分布式Hadoop(二)