add leetcode 1178 solution

leetcode/1178.find-Num-Of-Valid-Words/1178.go

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+package leetcode
+
+/*
+	匹配跟单词中的字母顺序，字母个数都无关，可以用bitmap压缩
+	1. 记录word中 利用map记录各种bit标示的个数
+	2. puzzles 中各个字母都不相同! 记录bitmap，然后搜索子空间中各种bit标识的个数的和
+		因为puzzles长度最长是7，所以搜索空间 2^7
+*/
+func findNumOfValidWords(words []string, puzzles []string) []int {
+
+	wordBitStatusMap := make(map[uint32]int, 0)
+	for _, w := range words {
+		wordBitStatusMap[toBitMap([]byte(w))]++
+	}
+	var res []int
+	for _, p := range puzzles {
+		var bitMap uint32
+		var totalNum int
+		bitMap |= (1 << (p[0] - 'a')) //work中要包含 p 的第一个字母 所以这个bit位上必须是1
+		findNum([]byte(p)[1:], bitMap, &totalNum, wordBitStatusMap)
+		res = append(res, totalNum)
+	}
+
+	return res
+}
+
+func toBitMap(word []byte) uint32 {
+	var res uint32
+	for _, b := range word {
+		res |= (1 << (b - 'a'))
+	}
+	return res
+}
+
+//利用dfs 搜索 pussles的子空间
+func findNum(puzzles []byte, bitMap uint32, totalNum *int, m map[uint32]int) {
+	if len(puzzles) == 0 {
+		*totalNum = *totalNum + m[bitMap]
+		return
+	}
+
+	//不包含puzzles[0],即puzzles[0]对应bit是0
+	findNum(puzzles[1:], bitMap, totalNum, m)
+
+	//包含puzzles[0],即puzzles[0]对应bit是1
+	bitMap |= (1 << (puzzles[0] - 'a'))
+	findNum(puzzles[1:], bitMap, totalNum, m)
+	bitMap ^= (1 << (puzzles[0] - 'a')) //异或 清零
+
+	return
+}

leetcode/1178.find-Num-Of-Valid-Words/1178_test.go

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+package leetcode
+
+import (
+	"reflect"
+	"testing"
+)
+
+func Test_findNumOfValidWords(t *testing.T) {
+
+	words1 := []string{"aaaa", "asas", "able", "ability", "actt", "actor", "access"}
+	puzzles1 := []string{"aboveyz", "abrodyz", "abslute", "absoryz", "actresz", "gaswxyz"}
+
+	type args struct {
+		words   []string
+		puzzles []string
+	}
+	tests := []struct {
+		name string
+		args args
+		want []int
+	}{
+		// TODO: Add test cases.
+		{"1", args{words: words1, puzzles: puzzles1}, []int{1, 1, 3, 2, 4, 0}},
+	}
+	for _, tt := range tests {
+		t.Run(tt.name, func(t *testing.T) {
+			if got := findNumOfValidWords(tt.args.words, tt.args.puzzles); !reflect.DeepEqual(got, tt.want) {
+				t.Errorf("findNumOfValidWords() = %v, want %v", got, tt.want)
+			}
+		})
+	}
+}

leetcode/1178.find-Num-Of-Valid-Words/README.md

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+
+
+# [1178. Number of Valid Words for Each Puzzle](https://leetcode-cn.com/problems/number-of-valid-words-for-each-puzzle/)
+
+
+## 题目
+
+With respect to a given puzzle string, a word is valid if both the following conditions are satisfied:
+word contains the first letter of puzzle.
+For each letter in word, that letter is in puzzle.
+For example, if the puzzle is "abcdefg", then valid words are "faced", "cabbage", and "baggage"; while invalid words are "beefed" (doesn't include "a") and "based" (includes "s" which isn't in the puzzle).
+Return an array answer, where answer[i] is the number of words in the given word list words that are valid with respect to the puzzle puzzles[i].
+
+**Example :**
+
+Input: 
+words = ["aaaa","asas","able","ability","actt","actor","access"], 
+puzzles = ["aboveyz","abrodyz","abslute","absoryz","actresz","gaswxyz"]
+Output: [1,1,3,2,4,0]
+Explanation:
+1 valid word for "aboveyz" : "aaaa" 
+1 valid word for "abrodyz" : "aaaa"
+3 valid words for "abslute" : "aaaa", "asas", "able"
+2 valid words for "absoryz" : "aaaa", "asas"
+4 valid words for "actresz" : "aaaa", "asas", "actt", "access"
+There're no valid words for "gaswxyz" cause none of the words in the list contains letter 'g'.
+
+**Constraints**:
+
+1 <= words.length <= 10^5
+4 <= words[i].length <= 50
+1 <= puzzles.length <= 10^4
+puzzles[i].length == 7
+words[i][j], puzzles[i][j] are English lowercase letters.
+Each puzzles[i] doesn't contain repeated characters.
+
+
+## 题目大意
+
+外国友人仿照中国字谜设计了一个英文版猜字谜小游戏，请你来猜猜看吧。
+
+字谜的迷面 puzzle 按字符串形式给出，如果一个单词 word 符合下面两个条件，那么它就可以算作谜底：
+
+单词 word 中包含谜面 puzzle 的第一个字母。
+单词 word 中的每一个字母都可以在谜面 puzzle 中找到。
+例如，如果字谜的谜面是 "abcdefg"，那么可以作为谜底的单词有 "faced", "cabbage", 和 "baggage"；而 "beefed"（不含字母 "a"）以及 "based"（其中的 "s" 没有出现在谜面中）都不能作为谜底。
+返回一个答案数组 answer，数组中的每个元素 answer[i] 是在给出的单词列表 words 中可以作为字谜迷面 puzzles[i] 所对应的谜底的单词数目。
+
+ 
+
+**示例：**
+
+输入：
+words = ["aaaa","asas","able","ability","actt","actor","access"], 
+puzzles = ["aboveyz","abrodyz","abslute","absoryz","actresz","gaswxyz"]
+输出：[1,1,3,2,4,0]
+**解释：**
+1 个单词可以作为 "aboveyz" 的谜底 : "aaaa" 
+1 个单词可以作为 "abrodyz" 的谜底 : "aaaa"
+3 个单词可以作为 "abslute" 的谜底 : "aaaa", "asas", "able"
+2 个单词可以作为 "absoryz" 的谜底 : "aaaa", "asas"
+4 个单词可以作为 "actresz" 的谜底 : "aaaa", "asas", "actt", "access"
+没有单词可以作为 "gaswxyz" 的谜底，因为列表中的单词都不含字母 'g'。
+
+**提示：**
+
+1 <= words.length <= 10^5
+4 <= words[i].length <= 50
+1 <= puzzles.length <= 10^4
+puzzles[i].length == 7
+words[i][j], puzzles[i][j] 都是小写英文字母。
+每个 puzzles[i] 所包含的字符都不重复。
+
+## 解题思路
+
+首先题目中两个限制条件非常关键： 
+
+- puzzles[i].length == 7
+- 每个 puzzles[i] 所包含的字符都不重复
+
+也就是说穷举每个puzzle的子串的搜索空间就是2^7=128，而且不用考虑去重问题。
+
+1. 因为谜底的判断只跟字符是否出现有关，跟字符的个数无关，另外都是小写的英文字母，所以可以用`bitmap`来表示单词(word)。
+2. 利用`map`记录不同状态的单词(word)的个数。
+3. 根据题意，如果某个单词(word)是某个字谜(puzzle)的谜底，那么word的bitmap肯定对应于puzzle某个子串的bitmap表示，且bitmap中包含puzzle的第一个字母的bit占用。
+4. 问题就转换为：求每一个puzzle的每一个子串，然后求和这个子串具有相同bitmap表示且word中包含puzzle的第一个字母的word的个数。