New Solution "Backspace String Compare"

Author: haoel

README.md

@@ -11,6 +11,7 @@ LeetCode
 |859|[Buddy Strings](https://leetcode.com/problems/buddy-strings/description/) | [C++](./algorithms/cpp/buddyStrings/BuddyStrings.cpp)|Easy|
 |858|[Mirror Reflection](https://leetcode.com/problems/mirror-reflection/description/) | [C++](./algorithms/cpp/mirrorReflection/MirrorReflection.cpp)|Medium|
 |852|[Peak Index in a Mountain Array](https://leetcode.com/problems/peak-index-in-a-mountain-array/description/) | [C++](./algorithms/cpp/peakIndexInAMountainArray/PeakIndexInAMountainArray.cpp)|Easy|
+844|[Backspace String Compare](https://leetcode.com/problems/backspace-string-compare/description/) | [C++](./algorithms/cpp/backspaceStringCompare/BackspaceStringCompare.cpp)|Easy|
 |837|[Most Common Word](https://leetcode.com/problems/most-common-word/) | [C++](./algorithms/cpp/mostCommonWord/MostCommonWord.cpp)|Easy|
 |804|[Unique Morse Code Words](https://leetcode.com/problems/unique-morse-code-words/description/) | [C++](./algorithms/cpp/uniqueMorseCodeWords/UniqueMorseCodeWords.cpp)|Easy|
 |771|[Jewels and Stones](https://leetcode.com/problems/jewels-and-stones/description) | [C++](./algorithms/cpp/jewelsAndStones/JewelsAndStones.cpp)|Easy|

algorithms/cpp/backspaceStringCompare/BackspaceStringCompare.cpp

@@ -0,0 +1,83 @@
+// Source : https://leetcode.com/problems/backspace-string-compare/description/
+// Author : Hao Chen
+// Date   : 2018-06-29
+
+/*************************************************************************************** 
+ *
+ * Given two strings S and T, return if they are equal when both are typed into empty 
+ * text editors. # means a backspace character.
+ * 
+ * 
+ * Example 1:
+ * 
+ * 
+ * Input: S = "ab#c", T = "ad#c"
+ * Output: true
+ * Explanation: Both S and T become "ac".
+ * 
+ * 
+ * 
+ * Example 2:
+ * 
+ * 
+ * Input: S = "ab##", T = "c#d#"
+ * Output: true
+ * Explanation: Both S and T become "".
+ * 
+ * 
+ * 
+ * Example 3:
+ * 
+ * 
+ * Input: S = "a##c", T = "#a#c"
+ * Output: true
+ * Explanation: Both S and T become "c".
+ * 
+ * 
+ * 
+ * Example 4:
+ * 
+ * 
+ * Input: S = "a#c", T = "b"
+ * Output: false
+ * Explanation: S becomes "c" while T becomes "b".
+ * 
+ * 
+ * Note:
+ * 
+ * 
+ * 	1 <= S.length <= 200
+ * 	1 <= T.length <= 200
+ * 	S and T only contain lowercase letters and '#' characters.
+ * 
+ * 
+ * Follow up:
+ * 
+ * 
+ * 	Can you solve it in O(N) time and O(1) space?
+ * 
+ * 
+ * 
+ * 
+ ***************************************************************************************/
+
+class Solution {
+private:    
+    void removeBackspaces(string &s) {
+        int i = 0;
+        for(int i=0; i<s.size(); i++) {
+            if (s[i] == '#') {
+                int backSteps = i>0 ? 2 : 1;
+                s.erase(i-backSteps + 1, backSteps);
+                i -= backSteps;
+            } 
+        }
+    }
+    
+public:
+    bool backspaceCompare(string S, string T) {
+        removeBackspaces(S);
+        removeBackspaces(T);
+        return S == T;
+    }
+};