标签：遍历 build help rar fun java 循环 lan char
比较含退格的字符串。题意是给两个字符串，中间包含井字，井字的意思是需要退格，请判断两个字符是否相等。例子，
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".
这个题比较简单，比较直观的思路是用stack做。followup有可能是会问如何做到不用额外空间，这里需要用到two pointer的做法。遍历的时候，需要从input的尾部开始遍历到头部，遇到井号的时候就记录一个skip++，如果不是井号但是skip大于0的话，就再i--；当skip为0的时候，会跳出while循环，这时比较S和T当前的char是否一样，若不一样则return false。如果两者有一个先遍历完了，也return false。
stack实现
时间O(n)
空间O(n)
Java
1 class Solution { 2 public boolean backspaceCompare(String S, String T) { 3 return build(S).equals(build(T)); 4 } 5 6 public String build(String S) { 7 Stack<Character> stack = new Stack(); 8 for (char c : S.toCharArray()) { 9 if (c != ‘#‘) { 10 stack.push(c); 11 } else if (c == ‘#‘ && !stack.empty()) { 12 stack.pop(); 13 } 14 } 15 return String.valueOf(stack); 16 } 17 }
JavaScript
1 /** 2 * @param {string} S 3 * @param {string} T 4 * @return {boolean} 5 */ 6 var backspaceCompare = function(S, T) { 7 return helper(S) == helper(T); 8 }; 9 10 var helper = function(input) { 11 let stack = []; 12 for (let c of input) { 13 if (c !== ‘#‘) { 14 stack.push(c); 15 } else if (c == ‘#‘ && stack.length > 0) { 16 stack.pop(); 17 } 18 } 19 return stack.join(‘‘); 20 };
two pointer实现
时间O(n)
空间O(1)
Java
1 class Solution { 2 public boolean backspaceCompare(String S, String T) { 3 int i = S.length() - 1; 4 int j = T.length() - 1; 5 int skipS = 0; 6 int skipT = 0; 7 8 while (i >= 0 || j >= 0) { // While there may be chars in build(S) or build (T) 9 while (i >= 0) { // Find position of next possible char in build(S) 10 if (S.charAt(i) == ‘#‘) { 11 skipS++; 12 i--; 13 } else if (skipS > 0) { 14 skipS--; 15 i--; 16 } else { 17 break; 18 } 19 } 20 while (j >= 0) { // Find position of next possible char in build(T) 21 if (T.charAt(j) == ‘#‘) { 22 skipT++; 23 j--; 24 } else if (skipT > 0) { 25 skipT--; 26 j--; 27 } else { 28 break; 29 } 30 } 31 // If two actual characters are different 32 if (i >= 0 && j >= 0 && S.charAt(i) != T.charAt(j)) { 33 return false; 34 } 35 // If expecting to compare char vs nothing 36 if ((i >= 0) != (j >= 0)) { 37 return false; 38 } 39 i--; 40 j--; 41 } 42 return true; 43 } 44 }
JavaScript
1 /** 2 * @param {string} S 3 * @param {string} T 4 * @return {boolean} 5 */ 6 var backspaceCompare = function(S, T) { 7 let i = S.length - 1; 8 let j = T.length - 1; 9 let skipS = 0; 10 let skipT = 0; 11 while (i >= 0 || j >= 0) { 12 while (i >= 0) { 13 if (S.charAt(i) == ‘#‘) { 14 skipS++; 15 i--; 16 } else if (skipS > 0) { 17 skipS--; 18 i--; 19 } else { 20 break; 21 } 22 } 23 24 while (j >= 0) { 25 if (T.charAt(j) == ‘#‘) { 26 skipT++; 27 j--; 28 } else if (skipT > 0) { 29 skipT--; 30 j--; 31 } else { 32 break; 33 } 34 } 35 36 if (i >= 0 && j >= 0 && S.charAt(i) !== T.charAt(j)) { 37 return false; 38 } 39 if (i >= 0 !== j >= 0) { 40 return false; 41 } 42 i--; 43 j--; 44 } 45 return true; 46 };
[LeetCode] 844. Backspace String Compare
标签：遍历 build help rar fun java 循环 lan char
原文地址：https://www.cnblogs.com/aaronliu1991/p/12670862.html
