add 41. First Missing Positive 🍺

README.md

@@ -43,6 +43,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 38 |[Count and Say](https://leetcode.com/problems/count-and-say)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/38.%20Count%20and%20Say.md)|Easy|  |
 | 39 |[Combination Sum](https://leetcode.com/problems/combination-sum/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/39.%20Combination%20Sum.md)|Medium|  |
 | 40 |[Combination Sum II](https://leetcode.com/problems/combination-sum-ii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/40.%20Combination%20Sum%20II.md)|Medium|  |
+| 41 |[First Missing Positive](https://leetcode.com/problems/first-missing-positive/)|[C++](solutions/41.%20First%20Missing%20Positive.md)|Hard|  |
 | 43 |[Multiply Strings](https://leetcode.com/problems/multiply-strings/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/43.%20Multiply%20Strings.md)|Medium|  |
 | 46 |[Permutations](https://leetcode.com/problems/permutations/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/46.%20Permutations.md)|Medium|  |
 | 47 |[Permutations II](https://leetcode.com/problems/permutations-ii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/47.%20Permutations%20II.md)|Medium|  |

solutions/41. First Missing Positive.md

@@ -0,0 +1,61 @@
+# [41. First Missing Positive](https://leetcode.com/problems/first-missing-positive/)
+
+# 思路
+
+题目要求找到缺失的首个正数。要求时间复杂度O(n)空间复杂度O(1)。
+
+
+## 思路一、nums数组只读
+
+提示告诉我们先不考虑额外的空间，那我们就可以用一个大小为n+1的bool数组bucket，`bucket[i] = true`表示原数组中存在 i，所以我们只需要遍历一遍原数组来填充bucket数组然后再从前往后遍历bucket，一旦遇到`bucket[i] = false`那么返回i即可。
+
+时间复杂度O(n)空间复杂度O(n)，空间复杂度没有达到要求，但这种方法的好处是不用修改原数组，所以如果原数组是只读的那这种方法还是不错的。
+
+## 思路二
+
+我们考虑将思路二中的额外bucket数组用原数组代替，即在nums[i]存放i（实际代码中存放的是i+1）。为此我们可以从前往后遍历数组，一旦遇到
+```
+nums[i] > 0 && nums[i] < n && nums[i] != nums[nums[i] - 1]
+```
+那么就不断交换`nums[i]`和`nums[nums[i] - 1]`直到上述条件不满足。
+
+由于每一次swap都会使有一个元素落在最终位置，所以最多交换n次，所以时间复杂度为O(n)，没有使用额外的空间，所以空间复杂度O(1)。
+
+# C++
+## 思路一
+``` C++
+class Solution {
+public:
+    int firstMissingPositive(vector<int>& nums) {
+        int n = nums.size();
+        vector<bool>bucket(n + 1, false);
+        
+        for(int num: nums)
+            if(num > 0 && num <= n) bucket[num] = true;
+        
+        for(int i = 1; i <= n; i++)
+            if(!bucket[i]) return i;
+        
+        return n + 1;
+    }
+};
+```
+
+## 思路二
+```C++
+class Solution {
+public:
+    int firstMissingPositive(vector<int>& nums) {
+        int n = nums.size();
+        
+        for(int i = 0; i < n; i++){
+            while(nums[i] > 0 && nums[i] < n && nums[i] != nums[nums[i] - 1])
+                swap(nums[i], nums[nums[i] - 1]);
+        }
+        for(int i = 0; i < n; i++)
+            if(nums[i] != i + 1) return i + 1;
+        
+        return n + 1;
+    }
+};
+```