A variation to a classical DP: LCS.

class Solution {
public:
/**
* @param A an integer array
* @return A list of integers includes the index of
* the first number and the index of the last number
*/
vector<int> continuousSubarraySum(vector<int>& A) {
vector<int> ret; size_t len = A.size(); int curr = A[];
int sum = A[], csum = A[];
int s = , e = , ss = , ee = ;
for (int i = ; i < len; i++)
{
int nsum = csum + A[i];
if (A[i] > nsum)
{
ss = ee = i;
csum = A[i];
}
else
{
ee = i;
csum = nsum;
} if(csum > sum) // update global record
{
sum = nsum;
s = ss;
e = ee;
}
} ret.push_back(s);
ret.push_back(e);
return ret;
}
};

LintCode "Continuous Subarray Sum"的更多相关文章

  1. [LintCode] Continuous Subarray Sum II

    Given an integer array, find a continuous rotate subarray where the sum of numbers is the biggest. Y ...

  2. [LintCode] Continuous Subarray Sum 连续子数组之和

    Given an integer array, find a continuous subarray where the sum of numbers is the biggest. Your cod ...

  3. LintCode 402: Continuous Subarray Sum

    LintCode 402: Continuous Subarray Sum 题目描述 给定一个整数数组,请找出一个连续子数组,使得该子数组的和最大.输出答案时,请分别返回第一个数字和最后一个数字的下标 ...

  4. Continuous Subarray Sum II(LintCode)

    Continuous Subarray Sum II   Given an circular integer array (the next element of the last element i ...

  5. leetcode 560. Subarray Sum Equals K 、523. Continuous Subarray Sum、 325.Maximum Size Subarray Sum Equals k(lintcode 911)

    整体上3个题都是求subarray,都是同一个思想,通过累加,然后判断和目标k值之间的关系,然后查看之前子数组的累加和. map的存储:560题是存储的当前的累加和与个数 561题是存储的当前累加和的 ...

  6. Continuous Subarray Sum

    Given an integer array, find a continuous subarray where the sum of numbers is the biggest. Your cod ...

  7. [LeetCode] Continuous Subarray Sum 连续的子数组之和

    Given a list of non-negative numbers and a target integer k, write a function to check if the array ...

  8. [Swift]LeetCode523. 连续的子数组和 | Continuous Subarray Sum

    Given a list of non-negative numbers and a target integer k, write a function to check if the array ...

  9. Continuous Subarray Sum LT523

    Given a list of non-negative numbers and a target integer k, write a function to check if the array ...

随机推荐

  1. 递归神经网络之理解长短期记忆网络(LSTM NetWorks)(转载)

    递归神经网络 人类并不是每时每刻都从头开始思考.正如你阅读这篇文章的时候,你是在理解前面词语的基础上来理解每个词.你不会丢弃所有已知的信息而从头开始思考.你的思想具有持续性. 传统的神经网络不能做到这 ...

  2. kafka的推和拉的问题

    之前学习过这一问题,但是面试又被问道了.再次记录下 推还是拉? Kafka最初考虑的问题是,customer应该从brokes拉取消息还是brokers将消息推送到consumer,也就是pull还p ...

  3. GNU C 扩展(转)

    GNU CC 是一个功能非常强大的跨平台 C 编译器,它对 C 语言提供了很多扩展,这些扩展对优化.目标代码布局.更安全的检查等方面提供了很强的支持.这里对支持支持 GNU 扩展的 C 语言成为 GN ...

  4. 26 个 jQuery使用技巧

    1. 禁用右键点击(Disable right-click) $(document).ready(function(){ $(document).bind("contextmenu" ...

  5. jquery轻松操作CSS样式

    $(this).click(function(){  if($(this).hasClass(“zxx_fri_on”)){    $(this).removeClass(“zxx_fri_on”); ...

  6. spring beans源码解读之--Bean的注解(annotation)

    随着spring注解的引入,越来越多的开发者开始使用注解,这篇文章将对注解的机制进行串联式的讲解,不求深入透彻,但求串起spring beans注解的珍珠,展示给大家. 1. spring beans ...

  7. caffe: test code Check failed: K_ == new_K (768 vs. 1024) Input size incompatible with inner product parameters.

    I0327 20:24:22.966171 20521 net.cpp:849] Copying source layer drop7I0327 20:24:22.966179 20521 net.c ...

  8. CENTOS 下编译HTK

    在centenos下编译HTK碰到缺少libX11库,需要安装 libX11-dev libXext-dev libXtst-dev

  9. Opencv 3入门(毛星云)摘要

    第一章 环境搭建: 1.  环境变量path 添加  D:\Program Files\opencv\build\x86\vc11\bin 2.  VS在VC++项目中,属性管理器\属性. VC++目 ...

  10. WCF入门教程五[WCF的通信模式]

    一.概述 WCF在通信过程中有三种模式:请求与答复.单向.双工通信.以下我们一一介绍. 二.请求与答复模式 描述: 客户端发送请求,然后一直等待服务端的响应(异步调用除外),期间处于假死状态,直到服务 ...