添加 problem 1160

Algorithms/1160. Find Words That Can Be Formed by Characters/1160. Find Words That Can Be Formed by Characters.go

@@ -0,0 +1,24 @@
+package leetcode
+
+func countCharacters(words []string, chars string) int {
+	count, res := make([]int, 26), 0
+	for i := 0; i < len(chars); i++ {
+		count[chars[i]-'a']++
+	}
+	for _, w := range words {
+		if canBeFormed(w, count) {
+			res += len(w)
+		}
+	}
+	return res
+}
+func canBeFormed(w string, c []int) bool {
+	count := make([]int, 26)
+	for i := 0; i < len(w); i++ {
+		count[w[i]-'a']++
+		if count[w[i]-'a'] > c[w[i]-'a'] {
+			return false
+		}
+	}
+	return true
+}

Algorithms/1160. Find Words That Can Be Formed by Characters/1160. Find Words That Can Be Formed by Characters_test.go

@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1160 struct {
+	para1160
+	ans1160
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1160 struct {
+	words []string
+	chars string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1160 struct {
+	one int
+}
+
+func Test_Problem1160(t *testing.T) {
+
+	qs := []question1160{
+
+		question1160{
+			para1160{[]string{"cat", "bt", "hat", "tree"}, "atach"},
+			ans1160{6},
+		},
+
+		question1160{
+			para1160{[]string{"hello", "world", "leetcode"}, "welldonehoneyr"},
+			ans1160{10},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1160------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1160, q.para1160
+		fmt.Printf("【input】:%v       【output】:%v\n", p, countCharacters(p.words, p.chars))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/1160. Find Words That Can Be Formed by Characters/README.md

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+# [1160. Find Words That Can Be Formed by Characters](https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/)
+
+
+## 题目:
+
+You are given an array of strings `words` and a string `chars`.
+
+A string is *good* if it can be formed by characters from `chars` (each character can only be used once).
+
+Return the sum of lengths of all good strings in `words`.
+
+**Example 1:**
+
+    Input: words = ["cat","bt","hat","tree"], chars = "atach"
+    Output: 6
+    Explanation: 
+    The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6.
+
+**Example 2:**
+
+    Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr"
+    Output: 10
+    Explanation: 
+    The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10.
+
+**Note:**
+
+1. `1 <= words.length <= 1000`
+2. `1 <= words[i].length, chars.length <= 100`
+3. All strings contain lowercase English letters only.
+
+
+## 题目大意
+
+
+给你一份『词汇表』（字符串数组） words 和一张『字母表』（字符串） chars。假如你可以用 chars 中的『字母』（字符）拼写出 words 中的某个『单词』（字符串），那么我们就认为你掌握了这个单词。注意：每次拼写时，chars 中的每个字母都只能用一次。返回词汇表 words 中你掌握的所有单词的 长度之和。
+
+提示：
+
+1. 1 <= words.length <= 1000
+2. 1 <= words[i].length, chars.length <= 100
+3. 所有字符串中都仅包含小写英文字母
+
+
+
+## 解题思路
+
+- 给出一个字符串数组 `words` 和一个字符串 `chars`，要求输出 `chars` 中能构成 `words` 字符串的字符数总数。
+- 简单题。先分别统计 `words` 和 `chars` 里面字符的频次。然后针对 `words` 中每个 `word` 判断能够能由 `chars` 构成，如果能构成，最终结果加上这个 `word` 的长度。