316:去除重复字母

LeetCode/src/main/java/leetcode/editor/cn/RemoveDuplicateLetters.java

@@ -0,0 +1,82 @@
+//给你一个字符串 s ，请你去除字符串中重复的字母，使得每个字母只出现一次。需保证 返回结果的字典序最小（要求不能打乱其他字符的相对位置）。 
+//
+// 注意：该题与 1081 https://leetcode-cn.com/problems/smallest-subsequence-of-distinct
+//-characters 相同 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//输入：s = "bcabc"
+//输出："abc"
+// 
+//
+// 示例 2： 
+//
+// 
+//输入：s = "cbacdcbc"
+//输出："acdb" 
+//
+// 
+//
+// 提示： 
+//
+// 
+// 1 <= s.length <= 104 
+// s 由小写英文字母组成 
+// 
+// Related Topics 栈 贪心算法 字符串 
+// 👍 518 👎 0
+
+package leetcode.editor.cn;
+
+import java.util.ArrayList;
+import java.util.List;
+import java.util.Stack;
+
+//316:去除重复字母
+public class RemoveDuplicateLetters {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new RemoveDuplicateLetters().new Solution();
+        System.out.println(solution.removeDuplicateLetters("bcabc"));//abc
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public String removeDuplicateLetters(String s) {
+            boolean[] vis = new boolean[26];
+            int[] num = new int[26];
+            for (int i = 0; i < s.length(); i++) {
+                num[s.charAt(i) - 'a']++;
+            }
+            Stack<Character> stack = new Stack<>();
+            for (int i = 0; i < s.length(); i++) {
+                char ch = s.charAt(i);
+                if (!vis[ch - 'a']) {
+                    while (!stack.empty() && stack.peek() > ch) {
+                        if (num[stack.peek() - 'a'] > 0) {
+                            vis[stack.peek() - 'a'] = false;
+                            stack.pop();
+                        } else {
+                            break;
+                        }
+                    }
+                    vis[ch - 'a'] = true;
+                    stack.push(ch);
+                }
+                num[ch - 'a'] -= 1;
+            }
+            StringBuilder sBuilder = new StringBuilder();
+            while (!stack.empty()) {
+                sBuilder.insert(0, stack.pop());
+            }
+            s = sBuilder.toString();
+            return s;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}

LeetCode/src/main/java/leetcode/editor/cn/RemoveDuplicateLetters.md

@@ -0,0 +1,28 @@
+<p>给你一个字符串 <code>s</code> ，请你去除字符串中重复的字母，使得每个字母只出现一次。需保证 <strong>返回结果的字典序最小</strong>（要求不能打乱其他字符的相对位置）。</p>
+
+<p><strong>注意：</strong>该题与 1081 <a href="https://leetcode-cn.com/problems/smallest-subsequence-of-distinct-characters">https://leetcode-cn.com/problems/smallest-subsequence-of-distinct-characters</a> 相同</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong><code>s = "bcabc"</code>
+<strong>输出<code>：</code></strong><code>"abc"</code>
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong><code>s = "cbacdcbc"</code>
+<strong>输出：</strong><code>"acdb"</code></pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 <= s.length <= 10<sup>4</sup></code></li>
+	<li><code>s</code> 由小写英文字母组成</li>
+</ul>
+<div><div>Related Topics</div><div><li>栈</li><li>贪心算法</li><li>字符串</li></div></div>\n<div><li>👍 518</li><li>👎 0</li></div>

LeetCode/src/main/java/leetcode/editor/cn/doc/java-by-pluto1314-y7xn.md

@@ -1,60 +0,0 @@
-### 解题思路
-执行用时：2 ms, 在所有 Java 提交中击败了100.00%的用户
-内存消耗：39 MB, 在所有 Java 提交中击败了9.73%的用户
-
-### 代码
-
-```java
-/**
- * Definition for singly-linked list.
- * public class ListNode {
- *     int val;
- *     ListNode next;
- *     ListNode(int x) { val = x; }
- * }
- */
-class Solution {
-    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
-        ListNode res = new ListNode(0);
-        ListNode helper = res;
-        int temp = 0;
-        int up = 0;
-
-        while (l1 != null && l2 != null) {
-            int sum = l1.val + l2.val + up;
-            if ( sum > 9) {
-                temp = sum - 10;
-                up = 1;
-            } else {
-                temp = sum;
-                up = 0;
-            }
-            helper.next = new ListNode(temp);
-            helper = helper.next;
-            l1 = l1.next;
-            l2 = l2.next;
-        }
-
-        ListNode helper2 = l1 == null ? l2 : l1;
-        while (helper2 != null) {
-            int sum = helper2.val + up;
-            if ( sum > 9) {
-                temp = sum - 10;
-                up = 1;
-            } else {
-                temp = sum;
-                up = 0;
-            }
-            helper.next = new ListNode(temp);
-            helper = helper.next;
-            helper2 = helper2.next;
-        }
-
-        if (up == 1) {
-            helper.next = new ListNode(up);
-            helper = helper.next;
-        }
-        return res.next;
-    }
-}
-```