字符串去重, 字典序输出(栈)

Author: cls1991

README.md

@@ -5,7 +5,7 @@ leetcode刷题, 直接用leetcode的分类方式.
 -  字符串
 - [x] 链表 (完成时间: 2018-01-28) 
 -  哈希表
--  堆栈
+- [x] 堆栈 (完成时间: 2018-02-07)
 -  双指针
 - [x] 树 (完成时间: 2018-02-04)
 -  图

Stack/316_RemoveDuplicateLetters.py

@@ -0,0 +1,52 @@
+# coding: utf8
+
+
+"""
+    题目链接: https://leetcode.com/problems/remove-duplicate-letters/description.
+    题目描述:
+
+    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.
+
+"""
+
+
+class Solution(object):
+    def removeDuplicateLetters(self, s):
+        """
+        :type s: str
+        :rtype: str
+        """
+        if not s or len(s) < 2:
+            return s
+
+        vis = [0] * 26
+        cnt = [0] * 26
+        ans = []
+        for c in s:
+            cnt[ord(c) - 97] += 1
+        # 遍历, 若当前字符已入栈, 跳过
+        # 否则, 与栈顶字符比较, 若小于栈顶字符且cnt[c] > 0(证明后面还会出现该字符), 出栈, 同时访问标记置为0
+        # 当前字符入栈, 访问标记置为1
+        for c in s:
+            idx = ord(c) - 97
+            cnt[idx] -= 1
+            if vis[idx]:
+                continue
+            while ans and c < ans[-1] and cnt[ord(ans[-1]) - 97]:
+                vis[ord(ans[-1]) - 97] = 0
+                ans.pop()
+            ans.append(c)
+            vis[idx] = 1
+
+        return ''.join(ans)