https://leetcode.cn/problems/backspace-string-compare/
844. Backspace String Compare Given two strings s and t, return true if they are equal when both are typed into empty text editors. '#' means a backspace character.
Note that after backspacing an empty text, the text will continue empty.
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 = "b" Output: false Explanation: s becomes "c" while t becomes "b".
Constraints:
1 <= s.length, 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?
func backspaceCompare(s string, t string) bool {
m, n := len(s), len(t)
p, q := m-1, n-1
for p > -1 && q > -1 {
i := 0
for p-i > -1 && s[p-i] == '#' {
i++
}
j := 0
for q-j > -1 && t[q-j] == '#' {
j++
}
p -= 2 * i
q -= 2 * j
if p < 0 && q < 0 {
return true
}
if p > -1 && q > -1 && s[p] != t[q] {
return false
}
p--
q--
}
return true
}
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标签：return,修复,text,&&,Input,true,bug,string,指针 From： https://www.cnblogs.com/rsapaper/p/16889883.html