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Formatted question description: https://leetcode.ca/all/1047.html

1047. Remove All Adjacent Duplicates In String (Easy)

Given a string S of lowercase letters, a duplicate removal consists of choosing two adjacent and equal letters, and removing them.

We repeatedly make duplicate removals on S until we no longer can.

Return the final string after all such duplicate removals have been made.  It is guaranteed the answer is unique.

 

Example 1:

Input: "abbaca"
Output: "ca"
Explanation: 
For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal, and this is the only possible move.  The result of this move is that the string is "aaca", of which only "aa" is possible, so the final string is "ca".

 

Note:

1. 1 <= S.length <= 20000
2. S consists only of English lowercase letters.

Related Topics:
Stack

Similar Questions:

• Remove All Adjacent Duplicates in String II (Medium)

Solution 1.

• Java
• C++
• Python
• Go
• Javascript
• RenderScript

• class Solution {
      public String removeDuplicates(String S) {
          Stack<Character> stack = new Stack<Character>();
          int length = S.length();
          for (int i = length - 1; i >= 0; i--) {
              char c = S.charAt(i);
              if (stack.isEmpty() || stack.peek() != c)
                  stack.push(c);
              else
                  stack.pop();
          }
          StringBuffer sb = new StringBuffer();
          while (!stack.isEmpty())
              sb.append(stack.pop());
          return sb.toString();
      }
  }
  
  ############
  
  class Solution {
      public String removeDuplicates(String s) {
          StringBuilder sb = new StringBuilder();
          for (char c : s.toCharArray()) {
              if (sb.length() > 0 && sb.charAt(sb.length() - 1) == c) {
                  sb.deleteCharAt(sb.length() - 1);
              } else {
                  sb.append(c);
              }
          }
          return sb.toString();
      }
  }
  

• // OJ: https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      string removeDuplicates(string s) {
          string ans;
          for (char c : s) {
              if (ans.empty() || c != ans.back()) ans += c;
              else ans.pop_back();
          }
          return ans;
      }
  };
  

• class Solution:
      def removeDuplicates(self, s: str) -> str:
          stk = []
          for c in s:
              if stk and stk[-1] == c:
                  stk.pop()
              else:
                  stk.append(c)
          return ''.join(stk)
  
  ############
  
  class Solution(object):
      def removeDuplicates(self, S):
          """
          :type S: str
          :rtype: str
          """
          stack = []
          for s in S:
              if stack and s == stack[-1]:
                  stack.pop()
                  continue
              stack.append(s)
          return "".join(stack)
  

• func removeDuplicates(s string) string {
  	stk := []rune{}
  	for _, c := range s {
  		if len(stk) > 0 && stk[len(stk)-1] == c {
  			stk = stk[:len(stk)-1]
  		} else {
  			stk = append(stk, c)
  		}
  	}
  	return string(stk)
  }
  

• /**
   * @param {string} s
   * @return {string}
   */
  var removeDuplicates = function (s) {
      const stk = [];
      for (const c of s) {
          if (stk.length && stk[stk.length - 1] == c) {
              stk.pop();
          } else {
              stk.push(c);
          }
      }
      return stk.join('');
  };
  
  

• impl Solution {
      pub fn remove_duplicates(s: String) -> String {
          let mut stack = Vec::new();
          for c in s.chars() {
              if !stack.is_empty() && *stack.last().unwrap() == c {
                  stack.pop();
              } else {
                  stack.push(c);
              }
          }
          stack.into_iter().collect()
      }
  }
  
  

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