标签：count div cal new dup example ret mon rem
Given a string which contains only lowercase letters, remove duplicate letters so that every letter appear once and only once. You must make sure your result is the smallest in lexicographical order among all possible results.
Example:
Given "bcabc"
Return "abc"
Given "cbacdcbc"
Return "acdb"
The runtime is O(26 * n) = O(n).
public class Solution { // 形成字典 int［26］ //找到当前str中最小的值 或者为 cnt为1的char //recursive： remove min 前面所有的char 然后将s中所有的charAt(min) 的字符移除； public String removeDuplicateLetters(String s) { int[] count = new int[26]; int min = 0; for(int i = 0 ; i < s.length() ; i++){ count[s.charAt(i) - ‘a‘] ++; } for(int i = 0 ; i < s.length(); i++){ if(s.charAt(i) < s.charAt(min)) min = i; if(--count[s.charAt(i) - ‘a‘] == 0) break; } return s.length()==0 ? "" : s.charAt(min) + removeDuplicateLetters(s.substring(min+1).replaceAll(""+s.charAt(min), "")); } }
标签：count div cal new dup example ret mon rem
原文地址：http://www.cnblogs.com/joannacode/p/6132579.html
