<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Broken Wings]]></title><description><![CDATA[新浪微博 Geopelia  欢迎关注多多交流～]]></description><link>https://blog.youkuaiyun.com/yanglei040</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; yanglei040]]></copyright><item><title><![CDATA[n个数求第k大]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50762863</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50762863</guid><author>yanglei040</author><pubDate>Mon, 29 Feb 2016 00:39:16 +0800</pubDate><description><![CDATA[借用快排的思想，每次选取一个参考值，把区间分成两块，左边都小于参考值，右面都大于参考值（中间如果有剩下就是和参考值相同），然后根据k再去选择排左面还是排右面
def select(a,l,r,k):
	i = l
	j = r
	print 'bg',i,j,k
	key = a[(l+r)>>1]
	while i<=j:
		while a[i]<key: i += 1
		while a[]]></description><category></category></item><item><title><![CDATA[leetcode 15. 3Sum]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50757617</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50757617</guid><author>yanglei040</author><pubDate>Sat, 27 Feb 2016 23:48:51 +0800</pubDate><description><![CDATA[给一个数组，找出所有的不相同的三元组相加为0.先排序，然后循环最小值，剩下两个值用two point从两侧向中间跑一遍，求出所有结果即可，复杂度o(n^2)。注意处理下重复class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: Li]]></description><category></category></item><item><title><![CDATA[leetcode 14. Longest Common Prefix]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50753559</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50753559</guid><author>yanglei040</author><pubDate>Sat, 27 Feb 2016 01:12:39 +0800</pubDate><description><![CDATA[给n个字符串，求最长公公前缀。直接逐位扫判断就行。
class Solution(object):
    def longestCommonPrefix(self, strs):
        """
        :type strs: List[str]
        :rtype: str
        """
        lenS = len(strs)
        i]]></description><category></category></item><item><title><![CDATA[leetcode 13. Roman to Integer]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50753558</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50753558</guid><author>yanglei040</author><pubDate>Sat, 27 Feb 2016 01:11:39 +0800</pubDate><description><![CDATA[给一个罗马数字，转化成整数。这个就简单点了，直接从前往后扫，只需要注意一个规则，如果小单位在大单位前，小单位位负，小单位在大单位后，小单位为正。
class Solution(object):
    def cov(self,s):
        if s == 'I': return 1
        elif s == 'V': return 5
        elif s == ']]></description><category></category></item><item><title><![CDATA[leetcode 12. Integer to Roman]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50753557</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50753557</guid><author>yanglei040</author><pubDate>Sat, 27 Feb 2016 01:08:50 +0800</pubDate><description><![CDATA[给一个整数（1-3999），转化成落马数字。直接模拟吧，注意特盘4和9的情况。罗马数字的规则参考http://zhidao.baidu.com/link?url=aFXj4HeKQCT1gv38YE7jjHz2f0KUiGS2E2yoCMRV4c3KDa9HnWo1FOIJR6l1MthPHSsEKMLp08VfkhiZ7wh3KK
大致以下几条：

  (1)基本数字Ⅰ、X 、C 中的任何]]></description><category></category></item><item><title><![CDATA[leetcode 11.Container With Most Water]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50753549</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50753549</guid><author>yanglei040</author><pubDate>Sat, 27 Feb 2016 01:06:46 +0800</pubDate><description><![CDATA[给一个数组a[]，整数对(i,ai)表示一个立在x轴上的隔板，求两个隔板使得隔板见可以存储最大容积的水。因为一对隔板存储的水量取决于低的板子，所以维护两个指针lp,rp如果a[lp]>a[rp],lp向后推的话得到的结果不可能比原来大，而rp向左推则可能会得到更大的答案，所以维护两个指针从两头向中间推一遍就好。
class Solution(object):
    def maxArea(se]]></description><category></category></item><item><title><![CDATA[leetcode 9. Palindrome Number]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50753543</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50753543</guid><author>yanglei040</author><pubDate>Sat, 27 Feb 2016 00:59:00 +0800</pubDate><description><![CDATA[判断一个数是否为回文数，直接循环判断
class Solution(object):
    def isPalindrome(self, x):
        """
        :type x: int
        :rtype: bool
        """
        s = str(x)
        lenS = len(s)
        
        f]]></description><category></category></item><item><title><![CDATA[leetcode 8. String to Integer (atoi)]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50753537</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50753537</guid><author>yanglei040</author><pubDate>Sat, 27 Feb 2016 00:58:02 +0800</pubDate><description><![CDATA[手写字符串转整数函数。细节好多，首先要处理掉前导空格，然后处理正负号，接下来往后处理时，处理到非数字字符时，立即结束返回前面得到的整数，最后如果溢出的话，返回离它最近的一个整数
class Solution(object):
    def myAtoi(self, str):
        """
        :type str: str
        :rtype: int
    ]]></description><category></category></item><item><title><![CDATA[leetcode 7. Reverse Integer]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50753536</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50753536</guid><author>yanglei040</author><pubDate>Sat, 27 Feb 2016 00:54:48 +0800</pubDate><description><![CDATA[给一个数，求各位翻转后的结果。还是模拟，处理下正负号就行
import math
class Solution(object):
    def reverse(self, x):
        """
        :type x: int
        :rtype: int
        """
        flag = 0
        if (x < 0) :
     ]]></description><category></category></item><item><title><![CDATA[leetcode 6. ZigZag Conversion]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50753478</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50753478</guid><author>yanglei040</author><pubDate>Sat, 27 Feb 2016 00:53:29 +0800</pubDate><description><![CDATA[给一串N型排列，把它转换成行排列。直接模拟，可以注意到，第一行和最后一行，间距都是2*step-2,第二行开始两个间距为2*step-4,2然后第一个间距依次＋2，第二个间距依次－2
class Solution(object):
    def convert(self, s, numRows):
        """
        :type s: str
        :type n]]></description><category></category></item><item><title><![CDATA[leetcode 5. Longest Palindromic Substring]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50743609</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50743609</guid><author>yanglei040</author><pubDate>Fri, 26 Feb 2016 00:28:43 +0800</pubDate><description><![CDATA[给一个字符串求最长公共字串。直接上o(n)的Manacher，方法网上很多讲的都挺详细，需要的自行百度吧
class Solution(object):
    def pre(self,s):
        lenS = len(s)
        i = 0
        res = []
        while i < lenS:
            res.append(']]></description><category></category></item><item><title><![CDATA[leetcode 3]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50734869</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50734869</guid><author>yanglei040</author><pubDate>Thu, 25 Feb 2016 01:35:18 +0800</pubDate><description><![CDATA[给一个字符串，求出最长的没有重复字符出现的字串。
two point直接推就好，右侧指针无法后移时更新一下答案
class Solution(object):
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        i ]]></description><category></category></item><item><title><![CDATA[leetcode 2]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50734865</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50734865</guid><author>yanglei040</author><pubDate>Thu, 25 Feb 2016 01:33:08 +0800</pubDate><description><![CDATA[给两个数值域小于10的单链表，按头对齐后对应位置相加，超过10要进位，够造成一个新的单链表。。。
直接相加，然后构造链表，处理下进位就行了
# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         sel]]></description><category></category></item><item><title><![CDATA[leetcode 1]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/50734860</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/50734860</guid><author>yanglei040</author><pubDate>Thu, 25 Feb 2016 01:29:19 +0800</pubDate><description><![CDATA[最近在学python,正好拿leetcode的题来练练基本语法也顺便补补代码量。。。好久没写代码了。。。
题意大概是给一个数组和一个数，求两个下标使得这两个下标的数之和等于目标数。
直接循环一个数，然后检查另一个数在不在数组里并且不能是第一个数本身，用py自带的函数解决就行了
class Solution(object):
    def twoSum(self, nums, target)]]></description><category></category></item><item><title><![CDATA[codefroces 486C Hack it!  数位dp+二分]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/39529065</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/39529065</guid><author>yanglei040</author><pubDate>Wed, 24 Sep 2014 21:10:59 +0800</pubDate><description><![CDATA[定义f(x)是x的各位数字之和(xshijinzhi)]]></description><category></category></item><item><title><![CDATA[hdu 5033 Building 几何+单调栈]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/39454203</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/39454203</guid><author>yanglei040</author><pubDate>Sun, 21 Sep 2014 19:35:52 +0800</pubDate><description><![CDATA[用一个单调栈存一下大楼的编号y]]></description><category></category></item><item><title><![CDATA[tc srm633 600 最大权闭合图]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/39401571</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/39401571</guid><author>yanglei040</author><pubDate>Fri, 19 Sep 2014 18:24:53 +0800</pubDate><description><![CDATA[给两棵树，两棵树编号相同的节点]]></description><category></category></item><item><title><![CDATA[poj2987 Firing 最大权闭合图 (最大流)]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/39379281</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/39379281</guid><author>yanglei040</author><pubDate>Thu, 18 Sep 2014 22:08:47 +0800</pubDate><description><![CDATA[题意是一个公司要裁员，每个员工有一个全]]></description><category></category></item><item><title><![CDATA[hdu5009 Paint Pearls  DP]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/39347283</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/39347283</guid><author>yanglei040</author><pubDate>Wed, 17 Sep 2014 20:21:05 +0800</pubDate><description><![CDATA[题意大概是给一个序列，每次可以取走连续的一段，花费是]]></description><category></category></item><item><title><![CDATA[hdu5017 Ellipsoid  三分套三分]]></title><link>https://blog.youkuaiyun.com/yanglei040/article/details/39343455</link><guid>https://blog.youkuaiyun.com/yanglei040/article/details/39343455</guid><author>yanglei040</author><pubDate>Wed, 17 Sep 2014 15:37:12 +0800</pubDate><description><![CDATA[题意大概是给一个椭球方程，求椭球面上到原点（0,0,0）的最小距离。
     三分x然后三分y,用]]></description><category></category></item></channel></rss>