feat: add solutions to lcof problem: No.66

Author: yanglbme

lcof/面试题38. 字符串的排列/Solution.py

@@ -15,4 +15,4 @@ def dfs(i):
         ans = []
         cs = list(s)
         dfs(0)
-        return ans
+        return ans

lcof/面试题57. 和为s的两个数字/Solution.js

@@ -16,4 +16,4 @@ var twoSum = function (nums, target) {
             ++l;
         }
     }
-};
+};

lcof/面试题66. 构建乘积数组/README.md

@@ -27,7 +27,17 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-`B[i] = (A[0] * A[1] * ... * A[i-1]) * (A[i+1] * ... * A[n-1])`
+**方法一：两次遍历**
+
+我们先创建一个长度为 $n$ 的答案数组 $ans$。
+
+接下来，我们从左到右遍历数组 $a$，过程中维护一个变量 $left$，表示当前元素左边所有元素的乘积，初始时 $left=1$。当遍历到 $a[i]$ 时，我们将 $left$ 赋值给 $ans[i]$，然后 $left$ 乘以 $a[i]$，即 $left \leftarrow left \times a[i]$。
+
+然后，我们从右到左遍历数组 $a$，过程中维护一个变量 $right$，表示当前元素右边所有元素的乘积，初始时 $right=1$。当遍历到 $a[i]$ 时，我们将 $ans[i]$ 乘上 $right$，然后 $right$ 乘以 $a[i]$，即 $right \leftarrow right \times a[i]$。
+
+最终，数组 $ans$ 即为所求的答案。
+
+时间复杂度 $O(n)$，其中 $n$ 为数组 $a$ 的长度。忽略答案数组的空间消耗，空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -39,15 +49,15 @@
 class Solution:
     def constructArr(self, a: List[int]) -> List[int]:
         n = len(a)
-        output = [1] * n
+        ans = [0] * n
         left = right = 1
         for i in range(n):
-            output[i] = left
+            ans[i] = left
             left *= a[i]
         for i in range(n - 1, -1, -1):
-            output[i] *= right
+            ans[i] *= right
             right *= a[i]
-        return output
+        return ans
 ```
 
 ### **Java**
@@ -58,17 +68,56 @@ class Solution:
 class Solution {
     public int[] constructArr(int[] a) {
         int n = a.length;
-        int[] output = new int[n];
+        int[] ans = new int[n];
         for (int i = 0, left = 1; i < n; ++i) {
-            output[i] = left;
+            ans[i] = left;
             left *= a[i];
         }
         for (int i = n - 1, right = 1; i >= 0; --i) {
-            output[i] *= right;
+            ans[i] *= right;
+            right *= a[i];
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> constructArr(vector<int>& a) {
+        int n = a.size();
+        vector<int> ans(n);
+        for (int i = 0, left = 1; i < n; ++i) {
+            ans[i] = left;
+            left *= a[i];
+        }
+        for (int i = n - 1, right = 1; ~i; --i) {
+            ans[i] *= right;
             right *= a[i];
         }
-        return output;
+        return ans;
     }
+};
+```
+
+### **Go**
+
+```go
+func constructArr(a []int) []int {
+	n := len(a)
+	ans := make([]int, n)
+	for i, left := 0, 1; i < n; i++ {
+		ans[i] = left
+		left *= a[i]
+	}
+	for i, right := n-1, 1; i >= 0; i-- {
+		ans[i] *= right
+		right *= a[i]
+	}
+	return ans
 }
 ```
 
@@ -81,16 +130,16 @@ class Solution {
  */
 var constructArr = function (a) {
     const n = a.length;
-    let output = new Array(n);
+    const ans = new Array(n);
     for (let i = 0, left = 1; i < n; ++i) {
-        output[i] = left;
+        ans[i] = left;
         left *= a[i];
     }
-    for (let i = n - 1, right = 1; i >= 0; --i) {
-        output[i] *= right;
+    for (let i = n - 1, right = 1; ~i; --i) {
+        ans[i] *= right;
         right *= a[i];
     }
-    return output;
+    return ans;
 };
 ```
 

lcof/面试题66. 构建乘积数组/Solution.cpp

@@ -0,0 +1,16 @@
+class Solution {
+public:
+    vector<int> constructArr(vector<int>& a) {
+        int n = a.size();
+        vector<int> ans(n);
+        for (int i = 0, left = 1; i < n; ++i) {
+            ans[i] = left;
+            left *= a[i];
+        }
+        for (int i = n - 1, right = 1; ~i; --i) {
+            ans[i] *= right;
+            right *= a[i];
+        }
+        return ans;
+    }
+};

lcof/面试题66. 构建乘积数组/Solution.go

@@ -0,0 +1,13 @@
+func constructArr(a []int) []int {
+	n := len(a)
+	ans := make([]int, n)
+	for i, left := 0, 1; i < n; i++ {
+		ans[i] = left
+		left *= a[i]
+	}
+	for i, right := n-1, 1; i >= 0; i-- {
+		ans[i] *= right
+		right *= a[i]
+	}
+	return ans
+}

lcof/面试题66. 构建乘积数组/Solution.java

@@ -1,15 +1,15 @@
 class Solution {
     public int[] constructArr(int[] a) {
         int n = a.length;
-        int[] output = new int[n];
+        int[] ans = new int[n];
         for (int i = 0, left = 1; i < n; ++i) {
-            output[i] = left;
+            ans[i] = left;
             left *= a[i];
         }
         for (int i = n - 1, right = 1; i >= 0; --i) {
-            output[i] *= right;
+            ans[i] *= right;
             right *= a[i];
         }
-        return output;
+        return ans;
     }
 }

lcof/面试题66. 构建乘积数组/Solution.js

@@ -4,14 +4,14 @@
  */
 var constructArr = function (a) {
     const n = a.length;
-    let output = new Array(n);
+    const ans = new Array(n);
     for (let i = 0, left = 1; i < n; ++i) {
-        output[i] = left;
+        ans[i] = left;
         left *= a[i];
     }
-    for (let i = n - 1, right = 1; i >= 0; --i) {
-        output[i] *= right;
+    for (let i = n - 1, right = 1; ~i; --i) {
+        ans[i] *= right;
         right *= a[i];
     }
-    return output;
+    return ans;
 };

lcof/面试题66. 构建乘积数组/Solution.py

@@ -1,12 +1,12 @@
 class Solution:
     def constructArr(self, a: List[int]) -> List[int]:
         n = len(a)
-        output = [1] * n
+        ans = [0] * n
         left = right = 1
         for i in range(n):
-            output[i] = left
+            ans[i] = left
             left *= a[i]
         for i in range(n - 1, -1, -1):
-            output[i] *= right
+            ans[i] *= right
             right *= a[i]
-        return output
+        return ans

solution/2500-2599/2557.Maximum Number of Integers to Choose From a Range II/README.md

@@ -61,7 +61,6 @@
 
 相似题目：[2557. 从一个范围内选择最多整数 II](/solution/2500-2599/2557.Maximum%20Number%20of%20Integers%20to%20Choose%20From%20a%20Range%20II/README.md)
 
-
 相似题目：[2556. 从一个范围内选择最多整数 I](/solution/2500-2599/2554.Maximum%20Number%20of%20Integers%20to%20Choose%20From%20a%20Range%20I/README.md)
 
 <!-- tabs:start -->

solution/2500-2599/2557.Maximum Number of Integers to Choose From a Range II/README_EN.md

@@ -43,7 +43,6 @@ All these integers are in the range [1, 7], all do not appear in banned, and the
 	<li><code>1 &lt;= maxSum &lt;= 10<sup>15</sup></code></li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->

solution/2500-2599/2559.Count Vowel Strings in Ranges/README_EN.md

@@ -8,7 +8,7 @@
 
 <p>Each query <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code> asks us to find the number of strings present in the range <code>l<sub>i</sub></code> to <code>r<sub>i</sub></code> (both <strong>inclusive</strong>) of <code>words</code> that start and end with a vowel.</p>
 
-<p>Return an array <code>ans</code> of size <code>queries.length</code>, where <code>ans[i]</code> is the answer to the <code>i</code><sup>th</sup> query.</p>
+<p>Return <em>an array </em><code>ans</code><em> of size </em><code>queries.length</code><em>, where </em><code>ans[i]</code><em> is the answer to the </em><code>i</code><sup>th</sup><em> query</em>.</p>
 
 <p><strong>Note</strong> that the vowel letters are <code>&#39;a&#39;</code>, <code>&#39;e&#39;</code>, <code>&#39;i&#39;</code>, <code>&#39;o&#39;</code>, and <code>&#39;u&#39;</code>.</p>
 

solution/2500-2599/2560.House Robber IV/README_EN.md

@@ -10,9 +10,9 @@
 
 <p>You are given an integer array <code>nums</code> representing how much money is stashed in each house. More formally, the <code>i<sup>th</sup></code> house from the left has <code>nums[i]</code> dollars.</p>
 
-<p>You are also given an integer <code>k</code>, representing the <strong>minimum</strong> number of houses the robber will steal from.&nbsp;It is always possible to steal at least <code>k</code> houses.</p>
+<p>You are also given an integer <code>k</code>, representing the <strong>minimum</strong> number of houses the robber will steal from. It is always possible to steal at least <code>k</code> houses.</p>
 
-<p>Return <em>the <strong>minimum</strong> </em>capability of the robber out of all the possible ways to steal at least <code>k</code> houses.</p>
+<p>Return <em>the <strong>minimum</strong> capability of the robber out of all the possible ways to steal at least </em><code>k</code><em> houses</em>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>