添加 problem 844

Algorithms/844. Backspace String Compare/844. Backspace String Compare.go

@@ -0,0 +1,25 @@
+package leetcode
+
+func backspaceCompare(S string, T string) bool {
+	s := make([]rune, 0)
+	for _, c := range S {
+		if c == '#' {
+			if len(s) > 0 {
+				s = s[:len(s)-1]
+			}
+		} else {
+			s = append(s, c)
+		}
+	}
+	s2 := make([]rune, 0)
+	for _, c := range T {
+		if c == '#' {
+			if len(s2) > 0 {
+				s2 = s2[:len(s2)-1]
+			}
+		} else {
+			s2 = append(s2, c)
+		}
+	}
+	return string(s) == string(s2)
+}

Algorithms/844. Backspace String Compare/844. Backspace String Compare_test.go

@@ -0,0 +1,58 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question844 struct {
+	para844
+	ans844
+}
+
+// para 是参数
+// one 代表第一个参数
+type para844 struct {
+	s string
+	t string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans844 struct {
+	one bool
+}
+
+func Test_Problem844(t *testing.T) {
+
+	qs := []question844{
+
+		question844{
+			para844{"ab#c", "ad#c"},
+			ans844{true},
+		},
+
+		question844{
+			para844{"ab##", "c#d#"},
+			ans844{true},
+		},
+
+		question844{
+			para844{"a##c", "#a#c"},
+			ans844{true},
+		},
+
+		question844{
+			para844{"a#c", "b"},
+			ans844{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 844------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans844, q.para844
+		fmt.Printf("【input】:%v       【output】:%v\n", p, backspaceCompare(p.s, p.t))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/844. Backspace String Compare/README.md

@@ -0,0 +1,68 @@
+# [844. Backspace String Compare](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:
+
+```c
+Input: S = "ab#c", T = "ad#c"
+Output: true
+Explanation: Both S and T become "ac".
+```
+
+Example 2:
+
+```c
+Input: S = "ab##", T = "c#d#"
+Output: true
+Explanation: Both S and T become "".
+```
+
+Example 3:
+
+```c
+Input: S = "a##c", T = "#a#c"
+Output: true
+Explanation: Both S and T become "c".
+```
+
+Example 4:
+
+```c
+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?
+
+## 题目大意
+
+
+给 2 个字符串，如果遇到 # 号字符，就回退一个字符。问最终的 2 个字符串是否完全一致。
+
+这一题可以用栈的思想来模拟，遇到 # 字符就回退一个字符。不是 # 号就入栈一个字符。比较最终 2 个字符串即可。
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