规范格式

leetcode/1002.Find-Common-Characters/1002. Find Common Characters.go

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+package leetcode
+
+import "math"
+
+func commonChars(A []string) []string {
+	cnt := [26]int{}
+	for i := range cnt {
+		cnt[i] = math.MaxUint16
+	}
+	cntInWord := [26]int{}
+	for _, word := range A {
+		for _, char := range []byte(word) { // compiler trick - here we will not allocate new memory
+			cntInWord[char-'a']++
+		}
+		for i := 0; i < 26; i++ {
+			// 缩小频次，使得统计的公共频次更加准确
+			if cntInWord[i] < cnt[i] {
+				cnt[i] = cntInWord[i]
+			}
+		}
+		// 重置状态
+		for i := range cntInWord {
+			cntInWord[i] = 0
+		}
+	}
+	result := make([]string, 0)
+	for i := 0; i < 26; i++ {
+		for j := 0; j < cnt[i]; j++ {
+			result = append(result, string(rune(i+'a')))
+		}
+	}
+	return result
+}

leetcode/1002.Find-Common-Characters/1002. Find Common Characters_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1002 struct {
+	para1002
+	ans1002
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1002 struct {
+	one []string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1002 struct {
+	one []string
+}
+
+func Test_Problem1002(t *testing.T) {
+
+	qs := []question1002{
+
+		question1002{
+			para1002{[]string{"bella", "label", "roller"}},
+			ans1002{[]string{"e", "l", "l"}},
+		},
+
+		question1002{
+			para1002{[]string{"cool", "lock", "cook"}},
+			ans1002{[]string{"c", "o"}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1002------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1002, q.para1002
+		fmt.Printf("【input】:%v       【output】:%v\n", p, commonChars(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1002.Find-Common-Characters/README.md

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+# [1002. Find Common Characters](https://leetcode.com/problems/find-common-characters/)
+
+
+## 题目:
+
+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.
+
+You may return the answer in any order.
+
+**Example 1:**
+
+    Input: ["bella","label","roller"]
+    Output: ["e","l","l"]
+
+**Example 2:**
+
+    Input: ["cool","lock","cook"]
+    Output: ["c","o"]
+
+**Note:**
+
+1. `1 <= A.length <= 100`
+2. `1 <= A[i].length <= 100`
+3. `A[i][j]` is a lowercase letter
+
+## 题目大意
+
+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。
+
+
+## 解题思路
+
+- 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。