标签:length amp string == bsp ring public 循环 ESS
记录一个current char, 用j循环到没有一样的为止,要是count>1的话,就把数字放进去
1 class Solution { 2 public int compress(char[] chars) { 3 if(chars.length == 0) return 0; 4 if(chars.length == 1) return 1; 5 6 int i = 0, j = 0; 7 int count = 0; 8 while(j < chars.length){ 9 char current = chars[j]; 10 while(j < chars.length && chars[j] == current){ 11 count++; 12 j++; 13 } 14 chars[i++] = current; 15 if(count != 1){ 16 String str = "" + count; 17 char[] arr = str.toCharArray(); 18 for(char c : arr){ 19 chars[i++] = c; 20 } 21 22 } 23 24 count = 0; 25 } 26 return i; 27 28 } 29 }
标签:length amp string == bsp ring public 循环 ESS
原文地址:https://www.cnblogs.com/goPanama/p/9747221.html