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"
题目意思：去掉重复的字符，保持元序列的位置，并使得字典序最小
真的是。。。想了半天也没想到用栈。看了他人的代码，还有用string 直接搞的
class Solution {
public:
string removeDuplicateLetters(string s) {
int n = s.length();
vector<int> cnt();
vector<int> vis();
for (int i = ; i < n; ++i) {
int x = s[i] - 'a';
cnt[x]++;
}
stack<int> st;
for (int i = ; i < n; ++i) {
int x = s[i] - 'a';
cnt[x]--;
if (st.empty()) {
st.push(x);
vis[x] = ;
}
else {
while (!st.empty() && st.top() > x && cnt[st.top()] && !vis[x]) {
vis[st.top()] = ;
st.pop();
}
if (!vis[x]) st.push(x),vis[x] = ;
}
}
string t = "";
while (!st.empty()) {
int y = st.top();
st.pop();
t = char(y + 'a') + t;
}
return t;
}
};class Solution {
public:
string removeDuplicateLetters(string s) {
vector<int> dict(, );
vector<bool> visited(, false);
for(auto ch : s) dict[ch]++;
string result = "";
/** the key idea is to keep a monotically increasing sequence **/
for(auto c : s) {
dict[c]--;
/** to filter the previously visited elements **/
if(visited[c]) continue;
while(c < result.back() && dict[result.back()]) {
visited[result.back()] = false;
result.pop_back();
}
result += c;
visited[c] = true;
}
return result.substr();
}
};