substring,subsequence,charAt执行效率的不同
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1 package com.java.tencent; 2 3 public class T_2_longestPalindrome { 4 5 public String test1(String s){ 6 long startTime=System.currentTimeMillis(); 7 int len = s.length(); 8 int tmp = 0; 9 String result = ""; 10 for(int i=0;i<len;i++){ 11 if(tmp+i-len>0){ 12 break; 13 } 14 int lenj = len-1; 15 for(int j=lenj;j>=i;j--){ 16 if(tmp>j-i+1){ 17 break; 18 } 19 String ch = s.substring(i,i+1); 20 String chj = s.substring(j,j+1); 21 if(ch.equals(chj)){ 22 String str = s.substring(i,j+1); 23 if(tmp>str.length()){ 24 continue; 25 } 26 boolean bl = true; 27 int ln = str.length(); 28 int lenm = ln/2+1; 29 for(int m=0;m<lenm;m++){ 30 CharSequence start = str.subSequence(m, m+1); 31 CharSequence end = str.subSequence(ln-m-1,ln-m); 32 if(!start.equals(end)){ 33 bl = false; 34 break; 35 } 36 } 37 if(bl && ln>tmp){ 38 result = str; 39 tmp = ln; 40 } 41 } 42 } 43 } 44 System.out.println(result); 45 long endTime=System.currentTimeMillis(); 46 System.out.println("程序运行时间: "+(endTime-startTime)+"ms"); 47 return result; 48 } 49 50 public String test2(String s){ 51 long startTime=System.currentTimeMillis(); 52 int len = s.length(); 53 int tmp = 0; 54 String result = ""; 55 for(int i=0;i<len;i++){ 56 if(tmp+i-len>0){ 57 break; 58 } 59 for(int j=len-1;j>=i;j--){ 60 if(tmp>j-i+1){ 61 break; 62 } 63 String str = s.substring(i,j+1); 64 boolean bl = true; 65 for(int m=0;m<(str.length()/2+1);m++){ 66 Character start = str.charAt(m); 67 Character end = str.charAt(str.length()-m-1); 68 if(!start.equals(end)){ 69 bl = false; 70 break; 71 } 72 } 73 if(bl && str.length()>tmp){ 74 result = str; 75 tmp = str.length(); 76 break; 77 } 78 } 79 } 80 System.out.println(result); 81 long endTime=System.currentTimeMillis(); 82 System.out.println("程序运行时间: "+(endTime-startTime)+"ms"); 83 return result; 84 } 85 86 87 public static void main(String[] args) { 88 T_2_longestPalindrome lp = new T_2_longestPalindrome(); 89 String s = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabcaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"; 90 lp.test2(s); 91 } 92 93 }
以上写了一个返回最长回文子串的程序(1000个a,回文字串就是如abcddcba、bcb,左右两边相同,当然a一个字符也是),之前使用的test1总是500以上的执行时间,时间总是无法降低,最后问题落到了两个截取字串位置上,经过修改,test2成功降低执行时间。
test1 使用 sub 截取字符串,执行时间500+ms
test2 使用 charAt,执行时间180ms
以上可以明显看出执行效率的不同。
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