problem:

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.

click to show more practice.

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.

Hide Tags

Divide and Conquer Array Dynamic
Programming

题意:找出一个序列的最大和连续子序列

thinking:

(1)这道题解法特别多:

方法1:将每个数和后一个数字相加,得到一个正负分布的序列,正数项对最大和子序列实用,作比較就可以

方法2:暴力匹配,两层循环,调用max()函数。时间复杂度O(n*n),也能够算出结果,可是提交超时

方法3:採用DP。时间复杂度O(n)

方法4:分治法。时间复杂度为nlog(n)

(2)本人实现了方法3 和方法4

code:

DP 法:

class Solution {
public:
int maxSubArray(int A[], int n) {
int sum=A[0];
int maxsum=A[0];
for(int i=1;i<n;i++)
{
if(sum<0) //DP核心
sum=0;
sum+=A[i];
maxsum=max(sum,maxsum);
}
return maxsum;
} };

分治法:

class Solution {
public:
int maxSubArray(int A[], int n) {
int ret=maxsub(A,0,n-1);
return ret; }
protected:
int maxsub(int A[], int start, int end)
{
int max_left=INT_MIN,max_mid=INT_MIN,max_right=INT_MIN;
if(start==end)
return A[start];
if(start+1==end)
{
int a=max(A[start],A[end]);
return a>(A[start]+A[end])?a:(A[start]+A[end]);
}
int mid=(start+end)/2;
int tmp_sum=A[mid];
max_mid=tmp_sum; int i=mid-1;
int j=mid+1;
while(i>=start) //难点在于当连续最大和子序列分布在mid的一側或两側时,怎么处理
{ tmp_sum+=A[i];
i--;
max_mid=max(max_mid,tmp_sum); }
if(max_mid>A[mid]) //推断是处于两側,还是处于一側
tmp_sum=max_mid;
else
tmp_sum=A[mid];
while(j<=end)
{ tmp_sum+=A[j];
j++;
max_mid=max(max_mid,tmp_sum); } max_left=max(max_left,maxsub(A,start,mid-1));//二分轮廓
max_right=max(max_right,maxsub(A,mid+1,end));
int tmp_max = max(max_left,max_right);
return max_mid>tmp_max?max_mid:tmp_max;
} };

leetcode || 53、Maximum Subarray的更多相关文章

  1. (LeetCode 53)Maximum Subarray

    Find the contiguous subarray within an array (containing at least one number) which has the largest ...

  2. 【LeetCode】053. Maximum Subarray

    题目: Find the contiguous subarray within an array (containing at least one number) which has the larg ...

  3. [LeetCode]题53:Maximum Subarray

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

  4. LeetCode(53) Maximum Subarray

    题目 Find the contiguous subarray within an array (containing at least one number) which has the large ...

  5. 【LeetCode算法-53】Maximum Subarray

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

  6. 【算法】LeetCode算法题-Maximum Subarray

    这是悦乐书的第154次更新,第156篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第13题(顺位题号是53).给定一个整数数组nums,找出一个最大和,此和是由数组中索引 ...

  7. 【leetcode】1186. Maximum Subarray Sum with One Deletion

    题目如下: Given an array of integers, return the maximum sum for a non-empty subarray (contiguous elemen ...

  8. LeetCode OJ:Maximum Subarray(子数组最大值)

    Find the contiguous subarray within an array (containing at least one number) which has the largest ...

  9. leetcode 53. Maximum Subarray 、152. Maximum Product Subarray

    53. Maximum Subarray 之前的值小于0就不加了.dp[i]表示以i结尾当前的最大和,所以需要用一个变量保存最大值. 动态规划的方法: class Solution { public: ...

随机推荐

  1. 构造函数后面的base()

    先执行父类的对应的构造函数,再执行当前的构造函数. 关于子类对象的构造函数和父类构造函数的执行顺序 以下内容转自:http://blog.csdn.net/todototry/article/deta ...

  2. java.lang.Throwable 异常/深入

    有几个现象是需要总结的: -------------------------------------- 在java语言中,错误类的基类是java.lang.Error,异常类的基类是java.lang ...

  3. ANDROID开发之SQLite详解

    本文转自:http://www.cnblogs.com/Excellent/archive/2011/11/19/2254888.html

  4. SQL 2005 日志损坏的恢复方法

    SQL 在突然停电或者非正常关机下,可能会出现日期文件错误,导致数据库不正常.恢复数据库方法如下 1.数据库服务停掉 将数据库文件备份 例如数据库名为 DTMS 则将 DTMS.mdf 备份出来. 2 ...

  5. java AES加密算法

    package com.siro.tools; import javax.crypto.Cipher;import javax.crypto.spec.IvParameterSpec;import j ...

  6. HDU 5122 K.Bro Sorting

    K.Bro Sorting Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) Tot ...

  7. memmove函数

    写一个函数,完成内存之间的拷贝 void* mymemcpy( void *dest, const void *src, size_t count ) { char* pdest = static_c ...

  8. 在Ubuntu下安装Apache

    在Ubuntu下安装软件其实非常方便,Ubuntu提供了apt-get工具,可以使用该工具直接下载安装软件. 在Linux里,系统最高权限账户为root账户,而默认登录的账户并非root账户,例如不具 ...

  9. HTTP协议学习笔记-2

    HTTP报文 HTTP报文分为请求报文和响应报文(requeset and response) 请求报文的格式: <method>  <request -URL> <ve ...

  10. Python常用模块的安装方法

    http://blog.163.com/yang_jianli/blog/static/161990006201162152724339/