add 204 find primes problem

Author: 6boris

src/0097.Interleaving-String/Solution.go

@@ -21,8 +21,7 @@ func isInterleave(s1 string, s2 string, s3 string) bool {
 			} else if j == 0 {
 				dp[i][j] = dp[i-1][j] && s1[i-1] == s3[i+j-1]
 			} else {
-				dp[i][j] = (dp[i-1][j] && s1[i-1] == s3[i+j-1]) ||
-					(dp[i][j-1] && s2[j-1] == s3[i+j-1])
+				dp[i][j] = (dp[i-1][j] && s1[i-1] == s3[i+j-1]) || (dp[i][j-1] && s2[j-1] == s3[i+j-1])
 			}
 		}
 	}

src/0204.Count-Primes/README.md

@@ -0,0 +1,73 @@
+# [204. Count Primes][title]
+
+## Description
+
+Count the number of prime numbers less than a non-negative number, n.
+
+**Example 1:**
+
+```
+Input: 10
+Output: 4
+Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.
+```
+
+**Tags:** Math, String
+
+## 题意
+> 求一个数以内的质数的个数
+
+## 题解
+
+### 思路1
+> 求质数可以用 `Sieve of Eratosthenes` 算法将时间复杂度降到
+
+```json
+（1）先把1删除（现今数学界1既不是质数也不是合数）
+（2）读取队列中当前最小的数2，然后把2的倍数删去
+（3）读取队列中当前最小的数3，然后把3的倍数删去
+（4）读取队列中当前最小的数5，然后把5的倍数删去
+（5）读取队列中当前最小的数7，然后把7的倍数删去
+（6）如上所述直到需求的范围内所有的数均删除或读取
+```
+
+```go
+func countPrimes(num int) int {
+	isPrime := make([]bool, num)
+	//	初始化数组 true表示都是质数
+	for i := 0; i < num; i++ {
+		isPrime[i] = true
+	}
+	//	循环检查质数 i<sqrt(n)可以转换为i*i < n
+	//	因为提劲的进行标记所以时间复杂可以到O(log N)
+	for i := 2; i*i < num; i++ {
+		if isPrime[i] == false {
+			continue
+		}
+		for j := i * i; j < num; j = j + i {
+			isPrime[j] = false
+		}
+	}
+	ans := 0
+	for i := 2; i < num; i++ {
+		if isPrime[i] {
+			ans++
+		}
+	}
+	return ans
+}
+
+```
+
+### 思路2
+> 思路2
+```go
+
+```
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-golang-leetcode][me]
+
+[title]: https://leetcode.com/problems/count-primes/
+[me]: https://github.com/kylesliu/awesome-golang-leetcode

src/0204.Count-Primes/Solution.go

@@ -0,0 +1,65 @@
+package Solution
+
+//	整合简化版
+func countPrimes(n int) int {
+	var ans int
+	prime := make([]bool, n)
+	for i := 2; i < n; i++ {
+		if prime[i] == false {
+			ans++
+			for j := 2; j*i < n; j++ {
+				prime[j*i] = true
+			}
+		}
+
+	}
+	return ans
+}
+
+func countPrimes2(n int) int {
+	var ans int
+	for i := 2; i < n; i++ {
+		if isPrime(i) {
+			ans++
+		}
+
+	}
+	return ans
+}
+
+func isPrime(num int) bool {
+	if num <= 1 {
+		return false
+	}
+	for i := 2; i*i <= num; i++ {
+		if num%i == 0 {
+			return false
+		}
+	}
+	return true
+}
+
+func countPrimes3(num int) int {
+	isPrime := make([]bool, num)
+	//	初始化数组 true表示都是质数
+	for i := 0; i < num; i++ {
+		isPrime[i] = true
+	}
+	//	循环检查质数 i<sqrt(n)可以转换为i*i < n
+	//	因为提劲的进行标记所以时间复杂可以到O(log N)
+	for i := 2; i*i < num; i++ {
+		if isPrime[i] == false {
+			continue
+		}
+		for j := i * i; j < num; j = j + i {
+			isPrime[j] = false
+		}
+	}
+	ans := 0
+	for i := 2; i < num; i++ {
+		if isPrime[i] {
+			ans++
+		}
+	}
+	return ans
+}

src/0204.Count-Primes/Solution_test.go

@@ -0,0 +1,82 @@
+package Solution
+
+import (
+	"reflect"
+	"testing"
+)
+
+func TestSolution(t *testing.T) {
+	//	测试用例
+	cases := []struct {
+		name   string
+		inputs int
+		expect int
+	}{
+		{"TestCase 1", 10, 4},
+	}
+
+	//	开始测试
+	for _, c := range cases {
+		t.Run(c.name, func(t *testing.T) {
+			got := countPrimes(c.inputs)
+			if !reflect.DeepEqual(got, c.expect) {
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
+					c.expect, got, c.inputs)
+			}
+		})
+	}
+}
+
+func TestSolution2(t *testing.T) {
+	//	测试用例
+	cases := []struct {
+		name   string
+		inputs int
+		expect int
+	}{
+		{"TestCase 1", 10, 4},
+	}
+
+	//	开始测试
+	for _, c := range cases {
+		t.Run(c.name, func(t *testing.T) {
+			got := countPrimes2(c.inputs)
+			if !reflect.DeepEqual(got, c.expect) {
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
+					c.expect, got, c.inputs)
+			}
+		})
+	}
+}
+
+
+func TestSolution3(t *testing.T) {
+	//	测试用例
+	cases := []struct {
+		name   string
+		inputs int
+		expect int
+	}{
+		{"TestCase 1", 10, 4},
+	}
+
+	//	开始测试
+	for _, c := range cases {
+		t.Run(c.name, func(t *testing.T) {
+			got := countPrimes3(c.inputs)
+			if !reflect.DeepEqual(got, c.expect) {
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
+					c.expect, got, c.inputs)
+			}
+		})
+	}
+}
+//	压力测试
+func BenchmarkSolution(b *testing.B) {
+
+}
+
+//	使用案列
+func ExampleSolution() {
+
+}