Add solution 1239

leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/1239. Maximum Length of a Concatenated String with Unique Characters.go

@@ -0,0 +1,38 @@
+package leetcode
+
+import (
+	"math/bits"
+)
+
+func maxLength(arr []string) int {
+	c, res := []uint32{}, 0
+	for _, s := range arr {
+		var mask uint32
+		for _, c := range s {
+			mask = mask | 1<<(c-'a')
+		}
+		if len(s) != bits.OnesCount32(mask) { // 如果字符串本身带有重复的字符，需要排除
+			continue
+		}
+		c = append(c, mask)
+	}
+	dfs(c, 0, 0, &res)
+	return res
+}
+
+func dfs(c []uint32, index int, mask uint32, res *int) {
+	*res = max(*res, bits.OnesCount32(mask))
+	for i := index; i < len(c); i++ {
+		if mask&c[i] == 0 {
+			dfs(c, i+1, mask|c[i], res)
+		}
+	}
+	return
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}

leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/1239. Maximum Length of a Concatenated String with Unique Characters_test.go

@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1239 struct {
+	para1239
+	ans1239
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1239 struct {
+	arr []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1239 struct {
+	one int
+}
+
+func Test_Problem1239(t *testing.T) {
+
+	qs := []question1239{
+
+		{
+			para1239{[]string{"un", "iq", "ue"}},
+			ans1239{4},
+		},
+
+		{
+			para1239{[]string{"cha", "r", "act", "ers"}},
+			ans1239{6},
+		},
+
+		{
+			para1239{[]string{"abcdefghijklmnopqrstuvwxyz"}},
+			ans1239{26},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1239------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1239, q.para1239
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxLength(p.arr))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1239.Maximum-Length-of-a-Concatenated-String-with-Unique-Characters/README.md

@@ -0,0 +1,88 @@
+# [1239. Maximum Length of a Concatenated String with Unique Characters](https://leetcode.com/problems/maximum-length-of-a-concatenated-string-with-unique-characters/)
+
+## 题目
+
+Given an array of strings `arr`. String `s` is a concatenation of a sub-sequence of `arr` which have **unique characters**.
+
+Return *the maximum possible length* of `s`.
+
+**Example 1:**
+
+```
+Input: arr = ["un","iq","ue"]
+Output: 4
+Explanation: All possible concatenations are "","un","iq","ue","uniq" and "ique".
+Maximum length is 4.
+```
+
+**Example 2:**
+
+```
+Input: arr = ["cha","r","act","ers"]
+Output: 6
+Explanation: Possible solutions are "chaers" and "acters".
+```
+
+**Example 3:**
+
+```
+Input: arr = ["abcdefghijklmnopqrstuvwxyz"]
+Output: 26
+```
+
+**Constraints:**
+
+- `1 <= arr.length <= 16`
+- `1 <= arr[i].length <= 26`
+- `arr[i]` contains only lower case English letters.
+
+## 题目大意
+
+给定一个字符串数组 arr，字符串 s 是将 arr 某一子序列字符串连接所得的字符串，如果 s 中的每一个字符都只出现过一次，那么它就是一个可行解。请返回所有可行解 s 中最长长度。
+
+## 解题思路
+
+- 每个字符串数组可以想象为 26 位的 0101 二进制串。出现的字符对应的位上标记为 1，没有出现的字符对应的位上标记为 0 。如果一个字符串中包含重复的字符，那么它所有 1 的个数一定不等于字符串的长度。如果 2 个字符串每个字母都只出现了一次，那么它们俩对应的二进制串 mask 相互与运算的结果一定为 0 ，即 0，1 互补了。利用这个特点，深搜所有解，保存出最长可行解的长度即可。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math/bits"
+)
+
+func maxLength(arr []string) int {
+	c, res := []uint32{}, 0
+	for _, s := range arr {
+		var mask uint32
+		for _, c := range s {
+			mask = mask | 1<<(c-'a')
+		}
+		if len(s) != bits.OnesCount32(mask) { // 如果字符串本身带有重复的字符，需要排除
+			continue
+		}
+		c = append(c, mask)
+	}
+	dfs(c, 0, 0, &res)
+	return res
+}
+
+func dfs(c []uint32, index int, mask uint32, res *int) {
+	*res = max(*res, bits.OnesCount32(mask))
+	for i := index; i < len(c); i++ {
+		if mask&c[i] == 0 {
+			dfs(c, i+1, mask|c[i], res)
+		}
+	}
+	return
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```