feat: add solutions to lc problem: No.0279.Perfect Squares

Author: maolonglong

solution/0200-0299/0279.Perfect Squares/README.md

@@ -18,7 +18,7 @@
 
 <pre>
 <strong>输入：</strong>n = <code>12</code>
-<strong>输出：</strong>3 
+<strong>输出：</strong>3
 <strong>解释：</strong><code>12 = 4 + 4 + 4</code></pre>
 
 <p><strong>示例 2：</strong></p>
@@ -40,22 +40,75 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+动态规划，定义 `dp[i]` 表示和为 `i` 的完全平方数的最少数量。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def numSquares(self, n: int) -> int:
+        dp = [0 for i in range(n + 1)]
+        for i in range(1, n + 1):
+            j, mi = 1, 0x3f3f3f3f
+            while j * j <= i:
+                mi = min(mi, dp[i - j * j])
+                j += 1
+            dp[i] = mi + 1
+        return dp[n]
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int numSquares(int n) {
+        List<Integer> ans = new ArrayList<>();
+        ans.add(0);
+        while (ans.size() <= n) {
+            int m = ans.size(), val = Integer.MAX_VALUE;
+            for (int i = 1; i * i <= m; i++) {
+                val = Math.min(val, ans.get(m - i * i) + 1);
+            }
+            ans.add(val);
+        }
+        return ans.get(n);
+    }
+}
+```
 
+### **Go**
+
+```go
+/*
+ * @lc app=leetcode.cn id=279 lang=golang
+ * 动态规划的思路，状态转移方程：dp[n] = min(dp[n-1*1]+1, dp[n-2*2]+1, ..., dp[n-k*k]+1), ( 0< k*k <=n )
+ */
+func numSquares(n int) int {
+	if n <= 0 {
+		return 0
+	}
+	dp := make([]int, n+1) // 多申请了一份整形，使代码更容易理解, dp[n] 就是 n 的完全平方数的求解
+	for i := 1; i <= n; i++ {
+		dp[i] = i // 初始值 dp[n] 的最大值的解，也是最容易求的解
+		for j := 0; j*j <= i; j++ {
+			dp[i] = minInt(dp[i-j*j]+1, dp[i])
+		}
+	}
+	return dp[n]
+}
+
+func minInt(x, y int) int {
+	if x < y {
+		return x
+	}
+	return y
+}
 ```
 
 ### **...**

solution/0200-0299/0279.Perfect Squares/README_EN.md

@@ -35,18 +35,71 @@
 
 ## Solutions
 
+For dynamic programming, define `dp[i]` to represent the least number of perfect square numbers that sum to `i`.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+class Solution:
+    def numSquares(self, n: int) -> int:
+        dp = [0 for i in range(n + 1)]
+        for i in range(1, n + 1):
+            j, mi = 1, 0x3f3f3f3f
+            while j * j <= i:
+                mi = min(mi, dp[i - j * j])
+                j += 1
+            dp[i] = mi + 1
+        return dp[n]
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public int numSquares(int n) {
+        List<Integer> ans = new ArrayList<>();
+        ans.add(0);
+        while (ans.size() <= n) {
+            int m = ans.size(), val = Integer.MAX_VALUE;
+            for (int i = 1; i * i <= m; i++) {
+                val = Math.min(val, ans.get(m - i * i) + 1);
+            }
+            ans.add(val);
+        }
+        return ans.get(n);
+    }
+}
+```
 
+### **Go**
+
+```go
+/*
+ * @lc app=leetcode.cn id=279 lang=golang
+ * 动态规划的思路，状态转移方程：dp[n] = min(dp[n-1*1]+1, dp[n-2*2]+1, ..., dp[n-k*k]+1), ( 0< k*k <=n )
+ */
+func numSquares(n int) int {
+	if n <= 0 {
+		return 0
+	}
+	dp := make([]int, n+1) // 多申请了一份整形，使代码更容易理解, dp[n] 就是 n 的完全平方数的求解
+	for i := 1; i <= n; i++ {
+		dp[i] = i // 初始值 dp[n] 的最大值的解，也是最容易求的解
+		for j := 0; j*j <= i; j++ {
+			dp[i] = minInt(dp[i-j*j]+1, dp[i])
+		}
+	}
+	return dp[n]
+}
+
+func minInt(x, y int) int {
+	if x < y {
+		return x
+	}
+	return y
+}
 ```
 
 ### **...**

solution/0200-0299/0279.Perfect Squares/Solution.java

@@ -1,14 +1,14 @@
 class Solution {
-	public int numSquares(int n) {
-		List<Integer> ans = new ArrayList<>();
-		ans.add(0);
-		while (ans.size() <= n) {
-			int m = ans.size(), val = Integer.MAX_VALUE;
-			for (int i = 1; i * i <= m; i++) {
-				val = Math.min(val, ans.get(m - i * i) + 1);
-			}
-			ans.add(val);
-		}
-		return ans.get(n);
-	}
-}
+    public int numSquares(int n) {
+        List<Integer> ans = new ArrayList<>();
+        ans.add(0);
+        while (ans.size() <= n) {
+            int m = ans.size(), val = Integer.MAX_VALUE;
+            for (int i = 1; i * i <= m; i++) {
+                val = Math.min(val, ans.get(m - i * i) + 1);
+            }
+            ans.add(val);
+        }
+        return ans.get(n);
+    }
+}

solution/0200-0299/0279.Perfect Squares/Solution.py

@@ -0,0 +1,10 @@
+class Solution:
+    def numSquares(self, n: int) -> int:
+        dp = [0 for i in range(n + 1)]
+        for i in range(1, n + 1):
+            j, mi = 1, 0x3f3f3f3f
+            while j * j <= i:
+                mi = min(mi, dp[i - j * j])
+                j += 1
+            dp[i] = mi + 1
+        return dp[n]