https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/

public class Solution {

public int longestSubstring(String s, int k) {

// Find count of each char

HashMap mp = new HashMap();

Object intObj;

for (int i=0; i<s.length(); i++) {

char ch = s.charAt(i);

intObj = mp.remove(ch);

int st = 0;

if (intObj != null) {

st = (int) intObj;

}

st++;

mp.put(ch, st);

}

// prepre iterate secondly

int ret = 0;

int last = -1;

HashMap newMp = new HashMap();

// iterate secondly

for (int i=0; i<s.length(); i++) {

char ch = s.charAt(i);

int num = (int)mp.get(ch);

// pass if num fits

if (num >= k) {

intObj = newMp.get(ch);

int newNum = 0;

if (intObj != null) {

newNum = (int) intObj;

}

newNum++;

newMp.put(ch, newNum);

continue;

}

// handle if meets nofit char

Set filter = new HashSet();

Iterator iter = newMp.entrySet().iterator();

Map.Entry entry;

// check newMp and prepare filter

while (iter.hasNext()) {

entry = (Map.Entry) iter.next();

char cch = (char)entry.getKey();

int cnt = (int)entry.getValue();

if (cnt < k) {

filter.add(cch);

}

int allCnt = (int)mp.remove(cch);

allCnt -= cnt;

mp.put(cch, allCnt);

}

// Prune

if (filter.size() == newMp.size()) {

last = i;

newMp.clear();

continue;

}

// use filter to check each segment

HashMap fMp = new HashMap();

int newLast = last;

for (int j=last+1; j<=i; j++) {

char fch = ' ';

if (j < i) {

fch = s.charAt(j);

}

// need to check segment

if (j == i || filter.contains(fch)) {

iter = fMp.entrySet().iterator();

// check map of each segment

while (iter.hasNext()) {

entry = (Map.Entry) iter.next();

char ffch = (char)entry.getKey();

int fcnt = (int)entry.getValue();

// Prepare Prune by update newMp

int newCnt = (int)newMp.remove(ffch);

newCnt -= fcnt;

newMp.put(ffch, newCnt);

if (newCnt < k) {

filter.add(ffch);

}

}

newLast = j;

fMp.clear();

// Check Prune

if (filter.size() == newMp.size()) {

break;

}

}

// no need to check segment, pass

else {

intObj = fMp.get(fch);

int fNum = 0;

if (intObj != null) {

fNum = (int) intObj;

}

fNum++;

fMp.put(fch, fNum);

if (fNum >= k) {

iter = fMp.entrySet().iterator();

boolean isFit = true;

while (iter.hasNext()) {

entry = (Map.Entry) iter.next();

int cfcnt = (int)entry.getValue();

if (cfcnt < k) {

isFit = false;

break;

}

}

if (isFit) {

if (j-newLast > ret) {

ret = j-newLast;

}

}

}

}

}

newMp.clear();

last = i;

}

if (s.length()-last-1 > ret) {

ret = s.length()-last-1;

}

return ret;

}

}

下面那个有未考虑到的地方,比如:

aaabbcbdcc

3

下面得出0,其实应该是3.

public class Solution {
public int longestSubstring(String s, int k) {
// Find count of each char
HashMap mp = new HashMap();
Object intObj;
for (int i=0; i<s.length(); i++) {
char ch = s.charAt(i);
intObj = mp.remove(ch);
int st = 0;
if (intObj != null) {
st = (int) intObj;
}
st++;
mp.put(ch, st);
} // prepre iterate secondly
int ret = 0;
int last = -1;
HashMap newMp = new HashMap(); // iterate secondly
for (int i=0; i<s.length(); i++) {
char ch = s.charAt(i);
int num = (int)mp.get(ch); // pass if num fits
if (num >= k) {
intObj = newMp.get(ch);
int newNum = 0;
if (intObj != null) {
newNum = (int) intObj;
}
newNum++;
newMp.put(ch, newNum);
continue;
} // handle if meets nofit char
Set filter = new HashSet();
Iterator iter = newMp.entrySet().iterator();
Map.Entry entry; // check newMp and prepare filter
while (iter.hasNext()) {
entry = (Map.Entry) iter.next();
char cch = (char)entry.getKey();
int cnt = (int)entry.getValue(); if (cnt < k) {
filter.add(cch);
} int allCnt = (int)mp.remove(cch);
allCnt -= cnt;
mp.put(cch, allCnt);
} // Prune
if (filter.size() == newMp.size()) {
last = i;
newMp.clear();
continue;
} // use filter to check each segment
HashMap fMp = new HashMap();
int newLast = last;
for (int j=last+1; j<=i; j++) {
char fch = ' ';
if (j < i) {
fch = s.charAt(j);
} // need to check segment
if (j == i || filter.contains(fch)) {
iter = fMp.entrySet().iterator();
boolean fits = true; // check map of each segment
while (iter.hasNext()) {
entry = (Map.Entry) iter.next();
char ffch = (char)entry.getKey();
int fcnt = (int)entry.getValue(); if (fcnt < k) {
fits = false;
} // Prepare Prune by update newMp
int newCnt = (int)newMp.remove(ffch);
newCnt -= fcnt;
newMp.put(ffch, newCnt); if (newCnt < k) {
filter.add(ffch);
}
} // update final ret
if (fits) {
if (j-newLast-1 > ret) {
ret = j-newLast-1;
}
}
newLast = j;
fMp.clear(); // Check Prune
if (filter.size() == newMp.size()) {
break;
} }
// no need to check segment, pass
else {
intObj = fMp.get(fch);
int fNum = 0;
if (intObj != null) {
fNum = (int) intObj;
}
fNum++;
fMp.put(fch, fNum);
}
}
newMp.clear();
last = i; }
if (s.length()-last-1 > ret) {
ret = s.length()-last-1;
}
return ret;
}
} test case: "zzzzzzzzzzaaaaaaaaabbbbbbbbhbhbhbhbhbhbhicbcbcibcbccccccccccbbbbbbbbaaaaaaaaafffaahhhhhiaahiiiiiiiiifeeeeeeeeee"
10 expected return: 21

longest-substring-with-at-least-k-repeating-characters的更多相关文章

  1. 395. Longest Substring with At Least K Repeating Characters

    395. Longest Substring with At Least K Repeating Characters 我的思路是先扫描一遍,然后判断是否都满足,否则,不满足的字符一定不出现,可以作为 ...

  2. [LeetCode] Longest Substring with At Least K Repeating Characters 至少有K个重复字符的最长子字符串

    Find the length of the longest substring T of a given string (consists of lowercase letters only) su ...

  3. [LeetCode] 395. Longest Substring with At Least K Repeating Characters 至少有K个重复字符的最长子字符串

    Find the length of the longest substring T of a given string (consists of lowercase letters only) su ...

  4. [Swift]LeetCode395. 至少有K个重复字符的最长子串 | Longest Substring with At Least K Repeating Characters

    Find the length of the longest substring T of a given string (consists of lowercase letters only) su ...

  5. LeetCode 395. Longest Substring with At Least K Repeating Characters C#

    Find the length of the longest substring T of a given string (consists of lowercase letters only) su ...

  6. 2016/9/21 leetcode 解题笔记 395.Longest Substring with At Least K Repeating Characters

    Find the length of the longest substring T of a given string (consists of lowercase letters only) su ...

  7. Leetcode: Longest Substring with At Least K Repeating Characters

    Find the length of the longest substring T of a given string (consists of lowercase letters only) su ...

  8. leetcode 395. Longest Substring with At Least K Repeating Characters

    Find the length of the longest substring T of a given string (consists of lowercase letters only) su ...

  9. 【LeetCode】395. Longest Substring with At Least K Repeating Characters 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/longest- ...

  10. leetcode 395. Longest Substring with At Least K Repeating Characters(高质量题)

    只能说还是太菜,抄的网上大神的做法: idea: mask 的每一位代表该位字母够不够k次,够k次为0,不够为1 对于每一位将其视为起点,遍历至末尾,找到其最大满足子串T的下标max_idx,之后从m ...

随机推荐

  1. java对对象排序

    一.前言 有时我们需要对类按照类中的某一个属性(或者多个属性)来对类的对象进行排序,有两种方法可以实现,一种方法是类实现Comparable<T>接口,然后调用Collections.so ...

  2. 用html5实现的flappy-bird

    可能网上早就有几个flappy-bird的html5版本啦,到这个时候flappy-bird可能也没有之前那么火了,但是作为一个新手,自己思考,自己动手写一个flappy-bird的demo还是很有成 ...

  3. mysql关于数据库表的水平拆分和垂直拆分

    最初知道水平垂直分表的时候是刚参加工作不久的时候,知道了这个概念,但是公司用户量和数据量始终没上来,所以也没用到过,知道有一天到了一家新公司后,这些才被应用到实际开发中,这里我就大概说说关于水平和垂直 ...

  4. 2016-2017 ACM-ICPC Pacific Northwest Regional Contest (Div. 2) 题解

    [题目链接] A - Alphabet 最长公共子序列.保留最长公共子序列,剩余的删除或者补足即可. #include <bits/stdc++.h> using namespace st ...

  5. CSUOJ 1018 Avatar

    Description In the planet Pandora, Jake found an old encryption algorithm. The plaintext, key and ci ...

  6. Django框架(一)-Django初识

    Django初识 一.Web框架本质—自己实现Web框架 1.所有的Web应用本质上就是一个socket服务端,而用户的浏览器就是一个socket客户端 import socket sk = sock ...

  7. 某谷 P5159 WD与矩阵

    题面在这里 崴脚回家后的小休闲2333. 显然每一行的1的个数必须是偶数,这样可以归纳证明前i行异或出来的m位二进制数也有偶数个1,这样最后一行就有且仅有一种放法了. 于是ans = 2^((n-1) ...

  8. 【并查集&&带权并查集】BZOJ3296&&POJ1182

    bzoj1529[POI2005]ska Piggy banks [题目大意] n头奶牛m种语言,每种奶牛分别掌握一些语言.问至少再让奶牛多学多少种语言,才能使得它们能够直接或间接交流? [思路] ( ...

  9. python开发_platform_获取操作系统详细信息工具

    ''' python中,platform模块给我们提供了很多方法去获取操作系统的信息 如: import platform platform.platform() #获取操作系统名称及版本号,'Win ...

  10. 基于(Redis | Memcache)实现分布式互斥锁

    设计一个缓存系统,不得不要考虑的问题就是:缓存穿透.缓存击穿与失效时的雪崩效应. 缓存击穿 缓存穿透是指查询一个一定不存在的数据,由于缓存是不命中时被动写的,并且出于容错考虑,如果从存储层查不到数据则 ...