316. Remove Duplicate Letters

区间寻找

思路

找到每个字母出现的最后位置,把整个串化为若干个区间,
找到每个区间的最小值,然后把字符串中最小值删除

TIPS

  • 区间开始一定是最小值第一次出现的地方
  • 区间结束是该原来的end或者新的end

代码

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public static int findMinPos(HashMap<Character, Integer> map) {
int pos = Integer.MAX_VALUE;
for (char c : map.keySet()) {
int x = map.get(c);
if (x < pos) {
pos = x;
}
}
return pos;
}
/**
* 找到所有字母最后出现的位置,然后按照这些位置依次划分区间,找到每个区间中最小的字母
*
* @param s
* @return
*/
public String removeDuplicateLetters(String s) {
if (s == null || s.length() == 0) {
return "";
}
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
char[] c = s.toCharArray();
for (int i = 0; i < c.length; i++) {
map.put(c[i], i);
}
int start = 0, end = findMinPos(map);
char[] res = new char[map.size()];
char min = 'z' + 1;
for (int j = 0; j < res.length; j++) {
// 区间内部找最小值
for (int i = start; i <= end; i++) {
//这里一定是c[i] < min,无等号,这样才能找到一个min出现的位置
if (c[i] < min && map.containsKey(c[i])) {
min = c[i];
start = i + 1;
}
}
// 先remove掉min,后面才可以重新查找
res[j] = min;
map.remove(min);
if (min == c[end]) {
end = findMinPos(map);
}
// 重置min
min = 'z' + 1;
}
return new String(res);
}

reference

easy-to-understand-iterative-java-solution

思路

  • int[] count: 记录每个字母出现的次数
  • boolean[] b: 是否已经被记录过
  • Stack stack: 结果集
    当栈中栈顶元素比当前元素大,且元素没有被记录过,以后还会出现的,出栈顶
    即栈中元素就是结果集,比栈顶大的元素才可以进栈

代码

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public String removeDuplicateLetters(String s) {
int[] count = new int[26];
boolean[] b = new boolean[26];
Stack<Character> stack = new Stack<Character>();
char[] c = s.toCharArray();
for (int i = 0; i < c.length; i++) {
count[c[i]-'a']++;
}
for (int i = 0; i < c.length; i++) {
count[c[i]-'a']--;
while(!stack.isEmpty() && stack.peek() > c[i] && !b[c[i]-'a']&& count[stack.peek()-'a'] > 0){
b[stack.peek()-'a'] = false;
stack.pop();
}
if(!b[c[i]-'a']){
stack.push(c[i]);
b[c[i]-'a'] = true;
}
}
StringBuffer sb = new StringBuffer();
while(!stack.isEmpty()){
sb.insert(0,stack.pop());
}
return sb.toString();
}

reference

java-solution-using-stack-with-comments