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"
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
根据字符串大小排序：优先把最小的字符放在前面，这样生成的字符肯定最小。问题是如何找出可以放在第一的最小字符呢？
首先，找出s中出现的所有字符最后一次出现的位置，再从中找到最小的位置index１，从０到ｉｎｄｅｘ１中找到最小字符，把选出来的字符做好标记
我们可以选从０到ｉｎｄｅｘ中的任意一个字符。其他所有待选字符都还在index-end中，等待之后选。
然后在剩余待选字符中找到最后一次出现位置的最小值ｉｎｄｅｘ２，再从ｉｎｄｅｘ１+1 到ｉｎｄｅｘ２里找出最小字符，把选出来的字符做好标记
重复…
时间复杂度Ｏ（n)
public class Solution {
public String removeDuplicateLetters(String s) {
int[] index = new int[26];
StringBuilder sb = new StringBuilder();
for (int i = 0; i < s.length(); i++) {
index[s.charAt(i) - 'a'] = i + 1;
}
int start = 0;
int cur = findMinIndex(index); // Return the index of the characters' last occurance in the string
while (cur >= 0) {
int chosen = findMinChar(s, start, cur, index); // Find the index of the smallest char from 0 to 'cur' in the string
start = chosen + 1;
index[s.charAt(chosen) - 'a'] = 0;
sb.append(s.charAt(chosen));
cur = findMinIndex(index);
}
return sb.toString();
}
private int findMinIndex(int[] index) {
int min = Integer.MAX_VALUE;
for (int i = 0; i < 26; i++) {
if (index[i] > 0 && min > index[i]) {
min = index[i];
}
}
if (min == Integer.MAX_VALUE) {
return -1;
} else {
return min - 1;
}
}
private int findMinChar(String s, int start, int end, int[] index) {
int min = -1;
char c = '{';
for (int i = start; i <= end; i++) {
if (index[s.charAt(i) - 'a'] > 0 && c > s.charAt(i)) {
c = s.charAt(i);
min = i;
}
}
return min;
}
}