标签：rate complex text with pointer point lower Plan ble
原题链接在这里：https://leetcode.com/problems/backspace-string-compare/
题目：
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 <= 2001 <= T.length <= 200S and T only contain lowercase letters and ‘#‘ characters.Follow up:
O(N) time and O(1) space?题解：
Could easily do it with stack. The follow up is to use O(N) time and O(1) space.
We could iterate the string back to front. Accumlate he # and use it when it is no #.
Then check if two chars are the same, if yes, move both pointers.
If not, check if two pointers alreay come to -1.
Time Complexity: O(m+n). m = S.length(). n = T.length().
Space: O(1).
AC Java:
1 class Solution { 2 public boolean backspaceCompare(String S, String T) { 3 if(S == null || T == null){ 4 return S == T; 5 } 6 7 int m = S.length(); 8 int n = T.length(); 9 int i = m - 1; 10 int j = n - 1; 11 int cntS = 0; 12 int cntT = 0; 13 14 while(i >= 0 || j >= 0){ 15 while(i >= 0 && (S.charAt(i) == ‘#‘ || cntS > 0)){ 16 if(S.charAt(i) == ‘#‘){ 17 cntS++; 18 }else{ 19 cntS--; 20 } 21 22 i--; 23 } 24 25 while(j >= 0 && (T.charAt(j) == ‘#‘ || cntT > 0)){ 26 if(T.charAt(j) == ‘#‘){ 27 cntT++; 28 }else{ 29 cntT--; 30 } 31 32 j--; 33 } 34 35 if(i >= 0 && j >= 0 && S.charAt(i) == T.charAt(j)){ 36 i--; 37 j--; 38 }else{ 39 return i == -1 && j == -1; 40 } 41 } 42 43 return true; 44 } 45 }
LeetCode 844. Backspace String Compare
标签：rate complex text with pointer point lower Plan ble
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/12110685.html
