feat: add python and java solutions to leetcode problem: No.0487

Author: yanglbme

README_EN.md

@@ -7,8 +7,8 @@
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solution/0400-0499/0487.Max Consecutive Ones II/README.md

@@ -33,22 +33,107 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+用 `prefix[i]` 数组表示以 i 结尾往前累计的最大连续 1 的个数，`suffix[i]` 数组表示以 i 开头往后累计的最大连续 1 的个数。
+
+遍历 `nums` 数组每个为 0 的位置，则位置 i 的最大连续 1 的个数为 `1 + prefix[i-1] + suffix[i+1]`。
+
+当然，如果 `nums` 数组没有 0，即所有元素都是 1，那么结果即为 `nums` 数组的长度。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
+        n = len(nums)
+        prefix = [0] * n
+        suffix = [0] * n
+        res = 0
+        for i in range(n):
+            if i == 0:
+                prefix[i] = nums[i]
+            else:
+                prefix[i] = 0 if nums[i] == 0 else prefix[i - 1] + 1
+            res = max(res, prefix[i])
+
+        for i in range(n - 1, -1, -1):
+            if i == n - 1:
+                suffix[i] = nums[i]
+            else:
+                suffix[i] = 0 if nums[i] == 0 else suffix[i + 1] + 1
+
+        for i in range(n):
+            if nums[i] == 0:
+                t = 1
+                if i > 0:
+                    t += prefix[i - 1]
+                if i < n - 1:
+                    t += suffix[i + 1]
+                res = max(res, t)
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
+- 双指针，时间复杂度 O(n²)，空间复杂度 O(1)
+
 ```java
+class Solution {
+    public int findMaxConsecutiveOnes(int[] nums) {
+        int n = nums.length;
+        int res = 0;
+        for (int i = 0; i < n; ++i) {
+            int cnt = 1;
+            int j = i;
+            while (j < n && (cnt > 0 || nums[j] == 1)) {
+                if (nums[j] == 0) --cnt;
+                ++j;
+            }
+            res = Math.max(res, j - i);
+        }
+        return res;
+    }
+}
+```
 
+- 辅助数组，时间复杂度 O(n)，空间复杂度 O(n)
+
+```java
+class Solution {
+    public int findMaxConsecutiveOnes(int[] nums) {
+        int n = nums.length;
+
+        int[] prefix = new int[n];
+        int[] suffix = new int[n];
+
+        int res = 0;
+        for (int i = 0; i < n; ++i) {
+            if (i == 0) prefix[0] = nums[0];
+            else prefix[i] = nums[i] == 0 ? 0 : prefix[i - 1] + 1;
+            res = Math.max(res, prefix[i]);
+        }
+
+        for (int i = n - 1; i >= 0; --i) {
+            if (i == n - 1) suffix[n - 1] = nums[n - 1];
+            else suffix[i] = nums[i] == 0 ? 0 : suffix[i + 1] + 1;
+        }
+
+        for (int i = 0; i < n; ++i) {
+            if (nums[i] == 0) {
+                int t = 1;
+                if (i > 0) t += prefix[i - 1];
+                if (i < n - 1) t += suffix[i + 1];
+                res = Math.max(res, t);
+            }
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/0400-0499/0487.Max Consecutive Ones II/README_EN.md

@@ -35,13 +35,92 @@ What if the input numbers come in one by one as an <b>infinite stream</b>? In ot
 ### **Python3**
 
 ```python
-
+class Solution:
+    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
+        n = len(nums)
+        prefix = [0] * n
+        suffix = [0] * n
+        res = 0
+        for i in range(n):
+            if i == 0:
+                prefix[i] = nums[i]
+            else:
+                prefix[i] = 0 if nums[i] == 0 else prefix[i - 1] + 1
+            res = max(res, prefix[i])
+
+        for i in range(n - 1, -1, -1):
+            if i == n - 1:
+                suffix[i] = nums[i]
+            else:
+                suffix[i] = 0 if nums[i] == 0 else suffix[i + 1] + 1
+
+        for i in range(n):
+            if nums[i] == 0:
+                t = 1
+                if i > 0:
+                    t += prefix[i - 1]
+                if i < n - 1:
+                    t += suffix[i + 1]
+                res = max(res, t)
+        return res
 ```
 
 ### **Java**
 
+- Two Pointers, time complexity: O(n²), memory complexity: O(1)
+
 ```java
+class Solution {
+    public int findMaxConsecutiveOnes(int[] nums) {
+        int n = nums.length;
+        int res = 0;
+        for (int i = 0; i < n; ++i) {
+            int cnt = 1;
+            int j = i;
+            while (j < n && (cnt > 0 || nums[j] == 1)) {
+                if (nums[j] == 0) --cnt;
+                ++j;
+            }
+            res = Math.max(res, j - i);
+        }
+        return res;
+    }
+}
+```
 
+- Prefix Array & Suffix Array, time complexity: O(n), memory complexity: O(n)
+
+```java
+class Solution {
+    public int findMaxConsecutiveOnes(int[] nums) {
+        int n = nums.length;
+
+        int[] prefix = new int[n];
+        int[] suffix = new int[n];
+
+        int res = 0;
+        for (int i = 0; i < n; ++i) {
+            if (i == 0) prefix[0] = nums[0];
+            else prefix[i] = nums[i] == 0 ? 0 : prefix[i - 1] + 1;
+            res = Math.max(res, prefix[i]);
+        }
+
+        for (int i = n - 1; i >= 0; --i) {
+            if (i == n - 1) suffix[n - 1] = nums[n - 1];
+            else suffix[i] = nums[i] == 0 ? 0 : suffix[i + 1] + 1;
+        }
+
+        for (int i = 0; i < n; ++i) {
+            if (nums[i] == 0) {
+                int t = 1;
+                if (i > 0) t += prefix[i - 1];
+                if (i < n - 1) t += suffix[i + 1];
+                res = Math.max(res, t);
+            }
+        }
+        return res;
+    }
+}
 ```
 
 ### **...**

solution/0400-0499/0487.Max Consecutive Ones II/Solution.java

@@ -0,0 +1,30 @@
+class Solution {
+    public int findMaxConsecutiveOnes(int[] nums) {
+        int n = nums.length;
+
+        int[] prefix = new int[n];
+        int[] suffix = new int[n];
+
+        int res = 0;
+        for (int i = 0; i < n; ++i) {
+            if (i == 0) prefix[0] = nums[0];
+            else prefix[i] = nums[i] == 0 ? 0 : prefix[i - 1] + 1;
+            res = Math.max(res, prefix[i]);
+        }
+
+        for (int i = n - 1; i >= 0; --i) {
+            if (i == n - 1) suffix[n - 1] = nums[n - 1];
+            else suffix[i] = nums[i] == 0 ? 0 : suffix[i + 1] + 1;
+        }
+
+        for (int i = 0; i < n; ++i) {
+            if (nums[i] == 0) {
+                int t = 1;
+                if (i > 0) t += prefix[i - 1];
+                if (i < n - 1) t += suffix[i + 1]; 
+                res = Math.max(res, t);
+            }
+        }
+        return res;
+    }
+}

solution/0400-0499/0487.Max Consecutive Ones II/Solution.py

@@ -0,0 +1,28 @@
+class Solution:
+    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
+        n = len(nums)
+        prefix = [0] * n
+        suffix = [0] * n
+        res = 0
+        for i in range(n):
+            if i == 0:
+                prefix[i] = nums[i]
+            else:
+                prefix[i] = 0 if nums[i] == 0 else prefix[i - 1] + 1
+            res = max(res, prefix[i])
+        
+        for i in range(n - 1, -1, -1):
+            if i == n - 1:
+                suffix[i] = nums[i]
+            else:
+                suffix[i] = 0 if nums[i] == 0 else suffix[i + 1] + 1
+        
+        for i in range(n):
+            if nums[i] == 0:
+                t = 1
+                if i > 0:
+                    t += prefix[i - 1]
+                if i < n - 1:
+                    t += suffix[i + 1]
+                res = max(res, t)
+        return res