规范格式

leetcode/0476.Number-Complement/476. Number Complement.go

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+package leetcode
+
+// 解法一
+func findComplement(num int) int {
+	xx := ^0 // ^0 = 1111111111111111111111
+	for xx&num > 0 {
+		xx <<= 1 // 构造出来的 xx = 1111111…000000，0 的个数就是 num 的长度
+	}
+	return ^xx ^ num // xx ^ num，结果是前面的 0 全是 1 的num，再取反即是答案
+}
+
+// 解法二
+func findComplement1(num int) int {
+	temp := 1
+	for temp <= num {
+		temp <<= 1 // 构造出来的 temp = 00000……10000，末尾 0 的个数是 num 的长度
+	}
+	return (temp - 1) ^ num // temp - 1 即是前面都是 0，num 长度的末尾都是 1 的数，再异或 num 即是最终结果
+}

leetcode/0476.Number-Complement/476. Number Complement_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question476 struct {
+	para476
+	ans476
+}
+
+// para 是参数
+// one 代表第一个参数
+type para476 struct {
+	one int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans476 struct {
+	one int
+}
+
+func Test_Problem476(t *testing.T) {
+
+	qs := []question476{
+
+		question476{
+			para476{5},
+			ans476{2},
+		},
+
+		question476{
+			para476{1},
+			ans476{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 476------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans476, q.para476
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findComplement(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0476.Number-Complement/README.md

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+# [476. Number Complement](https://leetcode.com/problems/number-complement/)
+
+
+## 题目:
+
+Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.
+
+**Note:**
+
+1. The given integer is guaranteed to fit within the range of a 32-bit signed integer.
+2. You could assume no leading zero bit in the integer’s binary representation.
+
+**Example 1:**
+
+    Input: 5
+    Output: 2
+    Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.
+
+**Example 2:**
+
+    Input: 1
+    Output: 0
+    Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.
+
+
+## 题目大意
+
+给定一个正整数，输出它的补数。补数是对该数的二进制表示取反。
+
+注意:
+
+给定的整数保证在32位带符号整数的范围内。
+你可以假定二进制数不包含前导零位。
+
+
+
+## 解题思路
+
+
+- 求一个正数的补数，补数的定义是对该数的二进制表示取反。当前不能改变符号位。按照题意构造相应的 mask 再取反即可。
+