feat: add solutions to lcof2 problems: No.004,005,006

Author: maolonglong

lcof2/剑指 Offer II 002. 二进制加法/README.md

@@ -41,7 +41,7 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-模拟笔算加法的过程，注意进位
+统计所有数字每个位中 1 出现的次数，对于某个位，1 出现的次数一定是 3 的倍数 +1 或 0。对这个数 %3 得到的结果就是那个出现一次的数字在该位上的值。
 
 <!-- tabs:start -->
 

lcof2/剑指 Offer II 004. 只出现一次的数字/README.md

@@ -45,22 +45,66 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+一位一位计算出结果
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def singleNumber(self, nums: List[int]) -> int:
+        ans = 0
+        for i in range(0, 32):
+            cnt = sum(num >> i & 1 for num in nums)
+            if cnt % 3:
+                if i == 31:
+                    ans -= 1 << i
+                else:
+                    ans |= 1 << i
+        return ans
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int singleNumber(int[] nums) {
+        int ans = 0;
+        for (int i = 0; i < 32; i++) {
+            int cnt = 0;
+            for (int num : nums) {
+                cnt += num >> i & 1;
+            }
+            cnt %= 3;
+            ans |= cnt << i;
+        }
+        return ans;
+    }
+}
+```
 
+### **Go**
+
+需要注意 Golang 中的 `int` 在 64 位平台上相当于 `int64`
+
+```go
+func singleNumber(nums []int) int {
+	ans := int32(0)
+	for i := 0; i < 32; i++ {
+		cnt := int32(0)
+		for _, num := range nums {
+			cnt += int32(num) >> i & 1
+		}
+		cnt %= 3
+		ans |= cnt << i
+	}
+	return int(ans)
+}
 ```
 
 ### **...**

lcof2/剑指 Offer II 004. 只出现一次的数字/Solution.go

@@ -0,0 +1,12 @@
+func singleNumber(nums []int) int {
+	ans := int32(0)
+	for i := 0; i < 32; i++ {
+		cnt := int32(0)
+		for _, num := range nums {
+			cnt += int32(num) >> i & 1
+		}
+		cnt %= 3
+		ans |= cnt << i
+	}
+	return int(ans)
+}

lcof2/剑指 Offer II 004. 只出现一次的数字/Solution.java

@@ -0,0 +1,14 @@
+class Solution {
+    public int singleNumber(int[] nums) {
+        int ans = 0;
+        for (int i = 0; i < 32; i++) {
+            int cnt = 0;
+            for (int num : nums) {
+                cnt += num >> i & 1;
+            }
+            cnt %= 3;
+            ans |= cnt << i;
+        }
+        return ans;
+    }
+}

lcof2/剑指 Offer II 004. 只出现一次的数字/Solution.py

@@ -0,0 +1,11 @@
+class Solution:
+    def singleNumber(self, nums: List[int]) -> int:
+        ans = 0
+        for i in range(0, 32):
+            cnt = sum(num >> i & 1 for num in nums)
+            if cnt % 3:
+                if i == 31:
+                    ans -= 1 << i
+                else:
+                    ans |= 1 << i
+        return ans

lcof2/剑指 Offer II 005. 单词长度的最大乘积/README.md

@@ -49,22 +49,85 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+因为只有 26 个小写字符，所以可以用一个 `int32` 存储字符的出现情况，然后枚举最大乘积
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def maxProduct(self, words: List[str]) -> int:
+        n = len(words)
+        mask = [0 for _ in range(n)]
+        for i, word in enumerate(words):
+            for ch in word:
+                mask[i] |= 1 << (ord(ch) - ord('a'))
+        ans = 0
+        for i in range(0, n - 1):
+            for j in range(i + 1, n):
+                if mask[i] & mask[j] == 0:
+                    ans = max(ans, len(words[i]) * len(words[j]))
+        return ans
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int maxProduct(String[] words) {
+        int n = words.length;
+        int[] mask = new int[n];
+        for (int i = 0; i < n; i++) {
+            for (char ch : words[i].toCharArray()) {
+                mask[i] |= 1 << (ch - 'a');
+            }
+        }
+        int ans = 0;
+        for (int i = 0; i < n - 1; i++) {
+            for (int j = i + 1; j < n; j++) {
+                if ((mask[i] & mask[j]) == 0) {
+                    ans = Math.max(ans, words[i].length() * words[j].length());
+                }
+            }
+        }
+        return ans;
+    }
+}
+```
 
+### **Go**
+
+```go
+func maxProduct(words []string) int {
+	n := len(words)
+	mask := make([]int32, n)
+	for i, word := range words {
+		for _, r := range word {
+			mask[i] |= 1 << (r - 'a')
+		}
+	}
+	ans := 0
+	for i := 0; i < n-1; i++ {
+		for j := i + 1; j < n; j++ {
+			if mask[i]&mask[j] == 0 {
+				ans = max(ans, len(words[i])*len(words[j]))
+			}
+		}
+	}
+	return ans
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
 ```
 
 ### **...**

lcof2/剑指 Offer II 005. 单词长度的最大乘积/Solution.go

@@ -0,0 +1,25 @@
+func maxProduct(words []string) int {
+	n := len(words)
+	mask := make([]int32, n)
+	for i, word := range words {
+		for _, r := range word {
+			mask[i] |= 1 << (r - 'a')
+		}
+	}
+	ans := 0
+	for i := 0; i < n-1; i++ {
+		for j := i + 1; j < n; j++ {
+			if mask[i]&mask[j] == 0 {
+				ans = max(ans, len(words[i])*len(words[j]))
+			}
+		}
+	}
+	return ans
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}

lcof2/剑指 Offer II 005. 单词长度的最大乘积/Solution.java

@@ -0,0 +1,20 @@
+class Solution {
+    public int maxProduct(String[] words) {
+        int n = words.length;
+        int[] mask = new int[n];
+        for (int i = 0; i < n; i++) {
+            for (char ch : words[i].toCharArray()) {
+                mask[i] |= 1 << (ch - 'a');
+            }
+        }
+        int ans = 0;
+        for (int i = 0; i < n - 1; i++) {
+            for (int j = i + 1; j < n; j++) {
+                if ((mask[i] & mask[j]) == 0) {
+                    ans = Math.max(ans, words[i].length() * words[j].length());
+                }
+            }
+        }
+        return ans;
+    }
+}

lcof2/剑指 Offer II 005. 单词长度的最大乘积/Solution.py

@@ -0,0 +1,13 @@
+class Solution:
+    def maxProduct(self, words: List[str]) -> int:
+        n = len(words)
+        mask = [0 for _ in range(n)]
+        for i, word in enumerate(words):
+            for ch in word:
+                mask[i] |= 1 << (ord(ch) - ord('a'))
+        ans = 0
+        for i in range(0, n - 1):
+            for j in range(i + 1, n):
+                if mask[i] & mask[j] == 0:
+                    ans = max(ans, len(words[i]) * len(words[j]))
+        return ans

lcof2/剑指 Offer II 006. 排序数组中两个数字之和/README.md

@@ -55,22 +55,62 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+双指针
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def twoSum(self, numbers: List[int], target: int) -> List[int]:
+        i, j = 0, len(numbers) - 1
+        while True:
+            if numbers[i] + numbers[j] < target:
+                i += 1
+            elif numbers[i] + numbers[j] > target:
+                j -= 1
+            else:
+                return [i, j]
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int[] twoSum(int[] numbers, int target) {
+        int i = 0, j = numbers.length - 1;
+        for (;;) {
+            if (numbers[i] + numbers[j] < target) {
+                i++;
+            } else if (numbers[i] + numbers[j] > target) {
+                j--;
+            } else {
+                return new int[]{i, j};
+            }
+        }
+    }
+}
+```
 
+### **Go**
+
+```go
+func twoSum(numbers []int, target int) []int {
+	for i, j := 0, len(numbers)-1; ; {
+		if numbers[i]+numbers[j] < target {
+			i++
+		} else if numbers[i]+numbers[j] > target {
+			j--
+		} else {
+			return []int{i, j}
+		}
+	}
+}
 ```
 
 ### **...**