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74 lines
2.4 KiB
Markdown
Executable File
74 lines
2.4 KiB
Markdown
Executable File
# [1002. Find Common Characters](https://leetcode.com/problems/find-common-characters/)
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## 题目
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Given an array `A` of strings made only from lowercase letters, return a list of all characters that show up in all strings within the list **(including duplicates)**. For example, if a character occurs 3 times in all strings but not 4 times, you need to include that character three times in the final answer.
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You may return the answer in any order.
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**Example 1**:
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Input: ["bella","label","roller"]
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Output: ["e","l","l"]
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**Example 2**:
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Input: ["cool","lock","cook"]
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Output: ["c","o"]
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**Note**:
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1. `1 <= A.length <= 100`
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2. `1 <= A[i].length <= 100`
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3. `A[i][j]` is a lowercase letter
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## 题目大意
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给定仅有小写字母组成的字符串数组 A，返回列表中的每个字符串中都显示的全部字符（包括重复字符）组成的列表。例如，如果一个字符在每个字符串中出现 3 次，但不是 4 次，则需要在最终答案中包含该字符 3 次。你可以按任意顺序返回答案。
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## 解题思路
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- 简单题。给出一个字符串数组 A，要求找出这个数组中每个字符串都包含字符，如果字符出现多次，在最终结果中也需要出现多次。这一题可以用 map 来统计每个字符串的频次，但是如果用数组统计会更快。题目中说了只有小写字母，那么用 2 个 26 位长度的数组就可以统计出来了。遍历字符串数组的过程中，不过的缩小每个字符在每个字符串中出现的频次(因为需要找所有字符串公共的字符，公共的频次肯定就是最小的频次)，得到了最终公共字符的频次数组以后，按顺序输出就可以了。
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## 代码
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```go
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package leetcode
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import "math"
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func commonChars(A []string) []string {
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cnt := [26]int{}
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for i := range cnt {
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cnt[i] = math.MaxUint16
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}
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cntInWord := [26]int{}
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for _, word := range A {
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for _, char := range []byte(word) { // compiler trick - here we will not allocate new memory
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cntInWord[char-'a']++
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}
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for i := 0; i < 26; i++ {
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// 缩小频次，使得统计的公共频次更加准确
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if cntInWord[i] < cnt[i] {
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cnt[i] = cntInWord[i]
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}
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}
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// 重置状态
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for i := range cntInWord {
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cntInWord[i] = 0
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}
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}
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result := make([]string, 0)
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for i := 0; i < 26; i++ {
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for j := 0; j < cnt[i]; j++ {
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result = append(result, string(i+'a'))
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}
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}
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return result
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}
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``` |