Merge pull request halfrost#195 from gostool/leetcode0859

Leetcode0859

Author: halfrost

leetcode/0859.Buddy-Strings/859.Buddy Strings.go

@@ -0,0 +1,30 @@
+package leetcode
+
+func buddyStrings(s string, goal string) bool {
+	if len(s) != len(goal) || len(s) <= 1 {
+		return false
+	}
+	mp := make(map[byte]int)
+	if s == goal {
+		for i := 0; i < len(s); i++ {
+			if _, ok := mp[s[i]]; ok {
+				return true
+			}
+			mp[s[i]]++
+		}
+		return false
+	}
+	first, second := -1, -1
+	for i := 0; i < len(s); i++ {
+		if s[i] != goal[i] {
+			if first == -1 {
+				first = i
+			} else if second == -1 {
+				second = i
+			} else {
+				return false
+			}
+		}
+	}
+	return second != -1 && s[first] == goal[second] && s[second] == goal[first]
+}

leetcode/0859.Buddy-Strings/859.Buddy Strings_test.go

@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question859 struct {
+	para859
+	ans859
+}
+
+// para 是参数
+type para859 struct {
+	s    string
+	goal string
+}
+
+// ans 是答案
+type ans859 struct {
+	ans bool
+}
+
+func Test_Problem859(t *testing.T) {
+
+	qs := []question859{
+
+		{
+			para859{"ab", "ba"},
+			ans859{true},
+		},
+
+		{
+			para859{"ab", "ab"},
+			ans859{false},
+		},
+
+		{
+			para859{"aa", "aa"},
+			ans859{true},
+		},
+
+		{
+			para859{"aaaaaaabc", "aaaaaaacb"},
+			ans859{true},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 859------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans859, q.para859
+		fmt.Printf("【input】:%v    【output】:%v\n", p, buddyStrings(p.s, p.goal))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0859.Buddy-Strings/README.md

@@ -0,0 +1,86 @@
+# [859. Buddy Strings](https://leetcode-cn.com/problems/buddy-strings/)
+
+## 题目
+
+Given two strings s and goal, return true if you can swap two letters in s so the result is equal to goal, otherwise, return false.
+
+Swapping letters is defined as taking two indices i and j (0-indexed) such that i != j and swapping the characters at s[i] and s[j].
+
+For example, swapping at indices 0 and 2 in "abcd" results in "cbad".
+
+**Example 1**:
+
+    Input: s = "ab", goal = "ba"
+    Output: true
+    Explanation: You can swap s[0] = 'a' and s[1] = 'b' to get "ba", which is equal to goal.
+
+**Example 2**:
+
+    Input: s = "ab", goal = "ab"
+    Output: false
+    Explanation: The only letters you can swap are s[0] = 'a' and s[1] = 'b', which results in "ba" != goal.
+
+**Example 3**:
+
+    Input: s = "aa", goal = "aa"
+    Output: true
+    Explanation: You can swap s[0] = 'a' and s[1] = 'a' to get "aa", which is equal to goal.
+
+**Example 4**:
+
+    Input: s = "aaaaaaabc", goal = "aaaaaaacb"
+    Output: true
+
+**Constraints:**
+
+- 1 <= s.length, goal.length <= 2 * 10000
+- s and goal consist of lowercase letters.
+
+## 题目大意
+
+给你两个字符串 s 和 goal ，只要我们可以通过交换 s 中的两个字母得到与 goal 相等的结果，就返回 true；否则返回 false 。
+
+交换字母的定义是：取两个下标 i 和 j （下标从 0 开始）且满足 i != j ，接着交换 s[i] 和 s[j] 处的字符。
+
+例如，在 "abcd" 中交换下标 0 和下标 2 的元素可以生成 "cbad" 。
+
+## 解题思路
+
+分为两种情况进行比较:
+- s等于goal,s中有重复元素就返回true,否则返回false
+- s不等于goal,s中有两个下标不同的字符与goal中对应下标的字符分别相等
+
+## 代码
+
+```go
+package leetcode
+
+func buddyStrings(s string, goal string) bool {
+	if len(s) != len(goal) || len(s) <= 1 {
+		return false
+	}
+	mp := make(map[byte]int)
+	if s == goal {
+		for i := 0; i < len(s); i++ {
+			if _, ok := mp[s[i]]; ok {
+				return true
+			}
+			mp[s[i]]++
+		}
+		return false
+	}
+	first, second := -1, -1
+	for i := 0; i < len(s); i++ {
+		if s[i] != goal[i] {
+			if first == -1 {
+				first = i
+			} else if second == -1 {
+				second = i
+			} else {
+				return false
+			}
+		}
+	}
+	return second != -1 && s[first] == goal[second] && s[second] == goal[first]
+}
+```