zoukankan      html  css  js  c++  java
  • Leetcode-Count and Say

    The count-and-say sequence is the sequence of integers beginning as follows:
    1, 11, 21, 1211, 111221, ...

    1 is read off as "one 1" or 11.
    11 is read off as "two 1s" or 21.
    21 is read off as "one 2, then one 1" or 1211.

    Given an integer n, generate the nth sequence.

    Note: The sequence of integers will be represented as a string.

    Analysis:

    The rule is:

    1 -> 11

    2 -> 12

    3 -> 13

    11 -> 21

    22 -> 22

    33 -> 23

    111 -> 31

    222 -> 32

    333 -> 33

    Solution:

     1 public class Solution {
     2     public String countAndSay(int n) {
     3         String curStr = "";
     4         if (n==0) return curStr;
     5         curStr="1";
     6         if (n==1) return curStr;
     7 
     8         for (int i=2;i<=n;i++){
     9             String newStr = "";
    10             int len = curStr.length();
    11             int index = 0;
    12             while (index<len){
    13                 char cur = curStr.charAt(index);
    14                 index++;
    15                 if (cur=='1') {
    16                     if (index+1<len && curStr.charAt(index)=='1' && curStr.charAt(index+1)=='1'){
    17                         newStr += "31";
    18                         index += 2;
    19                     } else if (index<len && curStr.charAt(index)=='1'){
    20                         newStr += "21";
    21                         index++;
    22                     } else newStr += "11";
    23                 } else if (cur=='2'){ 
    24                     if (index+1<len && curStr.charAt(index)=='2' && curStr.charAt(index+1)=='2'){
    25                         newStr += "32";
    26                         index += 2;
    27                     } else if (index<len && curStr.charAt(index)=='2'){
    28                         newStr += "22";
    29                         index++;
    30                     } else newStr += "12";
    31                 } else {
    32                     if (index+1<len && curStr.charAt(index)=='3' && curStr.charAt(index+1)=='3'){
    33                         newStr += "33";
    34                         index += 2;
    35                     } else if (index<len && curStr.charAt(index)=='3'){
    36                         newStr += "23";
    37                         index++;
    38                     } else newStr += "13";
    39                 }
    40             }
    41             curStr = newStr;
    42         }
    43 
    44         return curStr;       
    45     }
    46 }

     Solution 2:

    Solution 1 is a naive solution. Even though the number in the string cannot exceed 4, we should figure out a general solution.

     1 public class Solution {
     2     public String countAndSay(int n) {
     3         String res = "1";
     4         for (int i=2;i<=n;i++){            
     5             int index = 0;
     6             StringBuilder buf = new StringBuilder();
     7             while (index<res.length()){
     8                 int count = 1;
     9                 int index2 = index+1;
    10                 while (index2<res.length() && res.charAt(index2)==res.charAt(index)){
    11                     index2++;
    12                     count++;
    13                 }
    14 
    15                 buf.append((char)(count+'0'));
    16                 buf.append(res.charAt(index));
    17                 index = index2;
    18             }
    19             res = buf.toString();
    20         } 
    21 
    22         return res;
    23     }   
    24 }
  • 相关阅读:
    Java for LeetCode 229 Majority Element II
    Java for LeetCode 228 Summary Ranges
    Java for LeetCode 227 Basic Calculator II
    Java for LintCode 颜色分类
    Java for LintCode 链表插入排序
    Java for LintCode 颠倒整数
    Java for LintCode 验证二叉查找树
    Java for LeetCode 226 Invert Binary Tree
    Java for LeetCode 225 Implement Stack using Queues
    Java for LeetCode 224 Basic Calculator
  • 原文地址:https://www.cnblogs.com/lishiblog/p/4129937.html
Copyright © 2011-2022 走看看