我自己做出来的,分了几种情况来考虑。(再后面有加了注释的版本)

https://leetcode.com/problems/strong-password-checker/

// 加油!

public class Solution {

    public int strongPasswordChecker(String s) {
int sLen = s.length();
if (sLen < 4) {
return 6 - sLen;
}
int lnum = 1;
int unum = 1;
int dnum = 1;
int rcount = 0;
int ricount = 0;
int rdcount = 0;
int sameseq = 0; for (int i=0; i<sLen; i++) {
char ch = s.charAt(i);
if (ch>='a' && ch<='z') {
lnum = 0;
}
if (ch>='A' && ch<='Z') {
unum = 0;
}
if (ch>='0' && ch<='9') {
dnum = 0;
} // fix bug
if (i == 0) {
sameseq = 1;
}
else if (ch != s.charAt(i-1)) {
if (sameseq >= 3) {
// 这个很重要
while (sLen + ricount < 6 && sameseq >= 3) {
ricount++;
sameseq -= 2;
}
while (sLen - rdcount > 20 && sameseq >= 3) {
rdcount++;
sameseq --;
}
rcount += sameseq / 3;
}
sameseq = 1;
}
else {
sameseq++;
}
} // fixbug
if (sameseq >= 3) {
// 这个很重要
while (sLen + ricount < 6 && sameseq >= 3) {
ricount++;
sameseq -= 2;
}
while (sLen - rdcount > 20 && sameseq >= 3) {
rdcount++;
sameseq --;
}
rcount += sameseq / 3;
} //System.out.printf("rcount: %d, ricount: %d, rdcount: %d, lnum: %d, unum: %d, dnum: %d\n",
// rcount, ricount, rdcount, lnum, unum, dnum); int update = lnum + unum + dnum;
int must = ricount + rcount;
if (sLen + ricount < 6) {
must += 6 - sLen - ricount;
}
if (sLen < 20) {
return must > update ? must : update;
} // 跟上面的不一样,因为删除字符是无法增加新的类型的
if (sLen - rdcount > 20) {
rdcount += sLen - rdcount - 20;
}
return rcount >= update ? rcount + rdcount : update + rdcount; } }

以下是加了注释的版本:

public class Solution {

    public int strongPasswordChecker(String s) {
int sLen = s.length();
if (sLen < 4) {
return 6 - sLen;
} int lnum = 1; // need lower
int unum = 1; // need upper
int dnum = 1; // need digit int rcount = 0; // count need to replace repeated seq
int ricount = 0; // count need to add in repeated seq
int rdcount = 0; // count need to remove from repeated seq
int sameseq = 0; // count of chars in repeated seq for (int i=0; i<sLen; i++) {
char ch = s.charAt(i);
if (ch>='a' && ch<='z') {
lnum = 0;
}
if (ch>='A' && ch<='Z') {
unum = 0;
}
if (ch>='0' && ch<='9') {
dnum = 0;
} // check repeated seq
if (i == 0) {
sameseq = 1;
}
else if (ch != s.charAt(i-1)) {
if (sameseq >= 3) {
// if shorter length, add char into repeated seq
while (sLen + ricount < 6 && sameseq >= 3) {
ricount++;
sameseq -= 2;
}
// if longer length, remove char from repeated seq
while (sLen - rdcount > 20 && sameseq >= 3) {
rdcount++;
sameseq --;
}
// if length matches, replace char in repeated seq
rcount += sameseq / 3;
}
sameseq = 1;
}
else {
sameseq++;
}
} // need check repeated seq after loop
if (sameseq >= 3) {
// as previous process
while (sLen + ricount < 6 && sameseq >= 3) {
ricount++;
sameseq -= 2;
}
while (sLen - rdcount > 20 && sameseq >= 3) {
rdcount++;
sameseq --;
}
rcount += sameseq / 3;
} int update = lnum + unum + dnum;
int must = ricount + rcount;
if (sLen + ricount < 6) {
must += 6 - sLen - ricount;
}
if (sLen < 20) {
return must > update ? must : update;
} // if longer length, use below process
if (sLen - rdcount > 20) {
rdcount += sLen - rdcount - 20;
}
return rcount >= update ? rcount + rdcount : update + rdcount; } }

准备发表在Discuss版:

https://discuss.leetcode.com/category/549/strong-password-checker

【好】strong-password-checker,我自己做出来的:)的更多相关文章

  1. [LeetCode] Strong Password Checker 密码强度检查器

    A password is considered strong if below conditions are all met: It has at least 6 characters and at ...

  2. Leetcode: Strong Password Checker

    A password is considered strong if below conditions are all met: It has at least 6 characters and at ...

  3. [Swift]LeetCode420. 强密码检验器 | Strong Password Checker

    A password is considered strong if below conditions are all met: It has at least 6 characters and at ...

  4. Hard模式题目

    先过一下Hard模式的题目吧.   # Title Editorial Acceptance Difficulty Frequency   . 65 Valid Number     12.6% Ha ...

  5. 练练脑,继续过Hard题目

    http://www.cnblogs.com/charlesblc/p/6384132.html 继续过Hard模式的题目吧.   # Title Editorial Acceptance Diffi ...

  6. leetcode 学习心得 (2) (301~516)

    源代码地址:https://github.com/hopebo/hopelee 语言:C++ 301. Remove Invalid Parentheses Remove the minimum nu ...

  7. LeetCode All in One 题目讲解汇总(持续更新中...)

    终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 477 Total Hamming Distance ...

  8. Swift LeetCode 目录 | Catalog

    请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift    说明:题目中含有$符号则为付费题目. 如 ...

  9. LeetCode All in One题解汇总(持续更新中...)

    突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对 ...

  10. 继续过Hard题目

    接上一篇:http://www.cnblogs.com/charlesblc/p/6283064.html 继续过Hard模式的题目吧.   # Title Editorial Acceptance ...

随机推荐

  1. jstack 命令学习笔记

    大部分内容转载自:Java命令学习系列(二)--Jstack jstack - 查看堆栈信息 jstack ( Stack Trace for java ) 命令主要作用就是为了查看堆栈信息.它可以用 ...

  2. LeetCode解题报告—— Permutations & Permutations II & Rotate Image

    1. Permutations Given a collection of distinct numbers, return all possible permutations. For exampl ...

  3. python中的is, ==与对象的相等判断

    在java中,对于两个对象啊a,b,若a==b表示,a和b不仅值相等,而且指向同一内存位置,若仅仅比较值相等,应该用equals.而在python中对应上述两者的是‘is’ 和‘==’. (1) py ...

  4. netcore 配置文件使用

    一直在记录整理接口调用,但是最近发现关于项目在vs中本地启动也有许多方便的地方. 首先由于使用的是Java的Eureka和网关来做的服务基础, 然后服务就涉及到注册一说, 问题是,如果appsetti ...

  5. 前端自动化gulp遇上es6从 无知到深爱

    Gulp是什么? Gulp是前端自动化的工具,但Gulp能用来做什么 1.搭建web服务器 2.使用预处理器Sass,Less 3.压缩优化,可以压缩JS CSS Html 图片 4.自动将更新变化的 ...

  6. cuda8.0 百度云盘分享

    因为深度学习的需要,装了ubuntu16系统,同时也装了cuda,在下载cuda的时候发现教育网下载的速度不忍直视,故换了更快的网下载,结果发现10兆宽带下载速度依然很慢,不过总算还是下载了,故把千辛 ...

  7. AtCoder Regular Contest 103 Problem D Robot Arms (构造)

    题目链接  Problem D 给定$n$个坐标,然后让你构造一个长度为$m$的序列, 然后给每个坐标规定一个长度为$m$的序列,ULRD中的一个,意思是走的方向, 每次从原点出发按照这个序列方向,每 ...

  8. Single Number III(LintCode)

    Single Number III Given 2*n + 2 numbers, every numbers occurs twice except two, find them. Example G ...

  9. 洛谷——P2799 国王的魔镜

    P2799 国王的魔镜 题目描述 国王有一个魔镜,可以把任何接触镜面的东西变成原来的两倍——只是,因为是镜子嘛,增加的那部分是反的.比如一条项链,我们用AB来表示,不同的字母表示不同颜色的珍珠.如果把 ...

  10. Redux 和 Redux thunk 理解

    1: state 就像 model { todos: [{ text: 'Eat food', completed: true }, { text: 'Exercise', completed: fa ...