feat: add solutions to lc problem: No.0088

No.0088.Merge Sorted Array

Author: yanglbme

solution/0000-0099/0088.Merge Sorted Array/README.md

@@ -62,7 +62,13 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-双指针解决。
+**方法一：双指针**
+
+我们用两个指针 $i$ 和 $j$ 分别指向两个数组的末尾，用一个指针 $k$ 指向合并后的数组的末尾。
+
+每次比较两个数组的末尾元素，将较大的元素放在合并后的数组的末尾，然后将指针向前移动一位，重复这个过程，直到两个数组的指针都指向了数组的开头。
+
+时间复杂度 $O(m + n)$，其中 $m$ 和 $n$ 分别是两个数组的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -73,10 +79,8 @@
 ```python
 class Solution:
     def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
-        """
-        Do not return anything, modify nums1 in-place instead.
-        """
-        i, j, k = m - 1, n - 1, m + n - 1
+        k = m + n - 1
+        i, j = m - 1, n - 1
         while j >= 0:
             if i >= 0 and nums1[i] > nums2[j]:
                 nums1[k] = nums1[i]
@@ -87,56 +91,58 @@ class Solution:
             k -= 1
 ```
 
-```python
-class Solution:
-    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
-        """
-        Do not return anything, modify nums1 in-place instead.
-        """
-        i, j, k = m - 1, n - 1, m + n - 1
-        while j >= 0:
-            if i < 0 or nums1[i] < nums2[j]:
-                nums1[k] = nums2[j]
-                j -= 1
-            else:
-                nums1[k] = nums1[i]
-                i -= 1
-            k -= 1
-```
-
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
 class Solution {
     public void merge(int[] nums1, int m, int[] nums2, int n) {
-        int i = m - 1, j = n - 1, k = m + n - 1;
-        while (j >= 0) {
-            if (i >= 0 && nums1[i] > nums2[j]) {
-                nums1[k--] = nums1[i--];
-            } else {
-                nums1[k--] = nums2[j--];
-            }
+        for (int i = m - 1, j = n - 1, k = m + n - 1; j >= 0; --k) {
+            nums1[k] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
         }
     }
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
+        for (int i = m - 1, j = n - 1, k = m + n - 1; ~j; --k) {
+            nums1[k] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
+        }
+    }
+};
+```
+
 ### **Go**
 
 ```go
-func merge(nums1 []int, m int, nums2 []int, n int)  {
-    i, j, k := m - 1, n - 1, m + n - 1
-    for j >= 0 {
-        if i >= 0 && nums1[i] > nums2[j] {
-            nums1[k] = nums1[i]
-            i--
-        } else {
-            nums1[k] = nums2[j]
-            j--
-        }
-        k--
+func merge(nums1 []int, m int, nums2 []int, n int) {
+	for i, j, k := m-1, n-1, m+n-1; j >= 0; k-- {
+		if i >= 0 && nums1[i] > nums2[j] {
+			nums1[k] = nums1[i]
+			i--
+		} else {
+			nums1[k] = nums2[j]
+			j--
+		}
+	}
+}
+```
+
+### **TypeScript**
+
+```ts
+/**
+ Do not return anything, modify nums1 in-place instead.
+ */
+function merge(nums1: number[], m: number, nums2: number[], n: number): void {
+    for (let i = m - 1, j = n - 1, k = m + n - 1; j >= 0; --k) {
+        nums1[k] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
     }
 }
 ```
@@ -152,15 +158,8 @@ func merge(nums1 []int, m int, nums2 []int, n int)  {
  * @return {void} Do not return anything, modify nums1 in-place instead.
  */
 var merge = function (nums1, m, nums2, n) {
-    let i = m - 1,
-        j = n - 1,
-        k = m + n - 1;
-    while (j >= 0) {
-        if (i >= 0 && nums1[i] > nums2[j]) {
-            nums1[k--] = nums1[i--];
-        } else {
-            nums1[k--] = nums2[j--];
-        }
+    for (let i = m - 1, j = n - 1, k = m + n - 1; j >= 0; --k) {
+        nums1[k] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
     }
 };
 ```

solution/0000-0099/0088.Merge Sorted Array/README_EN.md

@@ -55,17 +55,23 @@ Note that because m = 0, there are no elements in nums1. The 0 is only there to
 
 ## Solutions
 
+**Approach 1: Two Pointers**
+
+We use two pointers $i$ and $j$ pointing to the end of two arrays, and a pointer $k$ pointing to the end of the merged array.
+
+Every time we compare the two elements at the end of the two arrays, and move the larger one to the end of the merged array. Then we move the pointer one step forward, and repeat this process until the two pointers reach the start of the arrays.
+
+The time complexity is $O(m + n)$, where $m$ and $n$ are the lengths of two arrays. The space complexity is $O(1)$.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
 class Solution:
     def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
-        """
-        Do not return anything, modify nums1 in-place instead.
-        """
-        i, j, k = m - 1, n - 1, m + n - 1
+        k = m + n - 1
+        i, j = m - 1, n - 1
         while j >= 0:
             if i >= 0 and nums1[i] > nums2[j]:
                 nums1[k] = nums1[i]
@@ -76,54 +82,56 @@ class Solution:
             k -= 1
 ```
 
-```python
-class Solution:
-    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
-        """
-        Do not return anything, modify nums1 in-place instead.
-        """
-        i, j, k = m - 1, n - 1, m + n - 1
-        while j >= 0:
-            if i < 0 or nums1[i] < nums2[j]:
-                nums1[k] = nums2[j]
-                j -= 1
-            else:
-                nums1[k] = nums1[i]
-                i -= 1
-            k -= 1
-```
-
 ### **Java**
 
 ```java
 class Solution {
     public void merge(int[] nums1, int m, int[] nums2, int n) {
-        int i = m - 1, j = n - 1, k = m + n - 1;
-        while (j >= 0) {
-            if (i >= 0 && nums1[i] > nums2[j]) {
-                nums1[k--] = nums1[i--];
-            } else {
-                nums1[k--] = nums2[j--];
-            }
+        for (int i = m - 1, j = n - 1, k = m + n - 1; j >= 0; --k) {
+            nums1[k] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
         }
     }
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
+        for (int i = m - 1, j = n - 1, k = m + n - 1; ~j; --k) {
+            nums1[k] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
+        }
+    }
+};
+```
+
 ### **Go**
 
 ```go
-func merge(nums1 []int, m int, nums2 []int, n int)  {
-    i, j, k := m - 1, n - 1, m + n - 1
-    for j >= 0 {
-        if i >= 0 && nums1[i] > nums2[j] {
-            nums1[k] = nums1[i]
-            i--
-        } else {
-            nums1[k] = nums2[j]
-            j--
-        }
-        k--
+func merge(nums1 []int, m int, nums2 []int, n int) {
+	for i, j, k := m-1, n-1, m+n-1; j >= 0; k-- {
+		if i >= 0 && nums1[i] > nums2[j] {
+			nums1[k] = nums1[i]
+			i--
+		} else {
+			nums1[k] = nums2[j]
+			j--
+		}
+	}
+}
+```
+
+### **TypeScript**
+
+```ts
+/**
+ Do not return anything, modify nums1 in-place instead.
+ */
+function merge(nums1: number[], m: number, nums2: number[], n: number): void {
+    for (let i = m - 1, j = n - 1, k = m + n - 1; j >= 0; --k) {
+        nums1[k] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
     }
 }
 ```
@@ -139,15 +147,8 @@ func merge(nums1 []int, m int, nums2 []int, n int)  {
  * @return {void} Do not return anything, modify nums1 in-place instead.
  */
 var merge = function (nums1, m, nums2, n) {
-    let i = m - 1,
-        j = n - 1,
-        k = m + n - 1;
-    while (j >= 0) {
-        if (i >= 0 && nums1[i] > nums2[j]) {
-            nums1[k--] = nums1[i--];
-        } else {
-            nums1[k--] = nums2[j--];
-        }
+    for (let i = m - 1, j = n - 1, k = m + n - 1; j >= 0; --k) {
+        nums1[k] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
     }
 };
 ```

solution/0000-0099/0088.Merge Sorted Array/Solution.cpp

@@ -0,0 +1,8 @@
+class Solution {
+public:
+    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
+        for (int i = m - 1, j = n - 1, k = m + n - 1; ~j; --k) {
+            nums1[k] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
+        }
+    }
+};

solution/0000-0099/0088.Merge Sorted Array/Solution.go

@@ -1,13 +1,11 @@
-func merge(nums1 []int, m int, nums2 []int, n int)  {
-    i, j, k := m - 1, n - 1, m + n - 1
-    for j >= 0 {
-        if i >= 0 && nums1[i] > nums2[j] {
-            nums1[k] = nums1[i]
-            i--
-        } else {
-            nums1[k] = nums2[j]
-            j--
-        }
-        k--
-    }
+func merge(nums1 []int, m int, nums2 []int, n int) {
+	for i, j, k := m-1, n-1, m+n-1; j >= 0; k-- {
+		if i >= 0 && nums1[i] > nums2[j] {
+			nums1[k] = nums1[i]
+			i--
+		} else {
+			nums1[k] = nums2[j]
+			j--
+		}
+	}
 }

solution/0000-0099/0088.Merge Sorted Array/Solution.java

@@ -1,12 +1,7 @@
 class Solution {
     public void merge(int[] nums1, int m, int[] nums2, int n) {
-        int i = m - 1, j = n - 1, k = m + n - 1;
-        while (j >= 0) {
-            if (i >= 0 && nums1[i] > nums2[j]) {
-                nums1[k--] = nums1[i--];
-            } else {
-                nums1[k--] = nums2[j--];
-            }
+        for (int i = m - 1, j = n - 1, k = m + n - 1; j >= 0; --k) {
+            nums1[k] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
         }
     }
 }

solution/0000-0099/0088.Merge Sorted Array/Solution.js

@@ -6,14 +6,7 @@
  * @return {void} Do not return anything, modify nums1 in-place instead.
  */
 var merge = function (nums1, m, nums2, n) {
-    let i = m - 1,
-        j = n - 1,
-        k = m + n - 1;
-    while (j >= 0) {
-        if (i >= 0 && nums1[i] > nums2[j]) {
-            nums1[k--] = nums1[i--];
-        } else {
-            nums1[k--] = nums2[j--];
-        }
+    for (let i = m - 1, j = n - 1, k = m + n - 1; j >= 0; --k) {
+        nums1[k] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
     }
 };

solution/0000-0099/0088.Merge Sorted Array/Solution.py

@@ -1,9 +1,7 @@
 class Solution:
     def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
-        """
-        Do not return anything, modify nums1 in-place instead.
-        """
-        i, j, k = m - 1, n - 1, m + n - 1
+        k = m + n - 1
+        i, j = m - 1, n - 1
         while j >= 0:
             if i >= 0 and nums1[i] > nums2[j]:
                 nums1[k] = nums1[i]

solution/0000-0099/0088.Merge Sorted Array/Solution.ts

@@ -0,0 +1,8 @@
+/**
+ Do not return anything, modify nums1 in-place instead.
+ */
+function merge(nums1: number[], m: number, nums2: number[], n: number): void {
+    for (let i = m - 1, j = n - 1, k = m + n - 1; j >= 0; --k) {
+        nums1[k] = i >= 0 && nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
+    }
+}