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docs/Leetcode_Solutions/C++/027._Remove_Element.md

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+# 27.Remove Element
+
+**<font color=red>难度Easy</font>**
+
+## 刷题内容
+> 原题连接
+
+* https://leetcode.com/problems/remove-element/
+
+> 内容描述
+
+```
+Given an array nums and a value val, remove all instances of that value in-place and return the new length.
+
+Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
+
+The order of elements can be changed. It doesn't matter what you leave beyond the new length.
+
+Example 1:
+
+Given nums = [3,2,2,3], val = 3,
+
+Your function should return length = 2, with the first two elements of nums being 2.
+
+It doesn't matter what you leave beyond the returned length.
+Example 2:
+
+Given nums = [0,1,2,2,3,0,4,2], val = 2,
+
+Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.
+
+Note that the order of those five elements can be arbitrary.
+
+It doesn't matter what values are set beyond the returned length.
+Clarification:
+
+Confused why the returned value is an integer but your answer is an array?
+
+Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
+
+Internally you can think of this:
+
+// nums is passed in by reference. (i.e., without making a copy)
+int len = removeElement(nums, val);
+
+// any modification to nums in your function would be known by the caller.
+// using the length returned by your function, it prints the first len elements.
+for (int i = 0; i < len; i++) {
+    print(nums[i]);
+}
+```
+> 思路
+******- 时间复杂度: O(n)******- 空间复杂度: O(1)******
+
+我们可以遍历数组，把等于 val 的数放到数组的后半部分就行。我们可以用双指针实现。当 nums[i] != val 时，nums[j++] = nums[i]
+```cpp
+class Solution {
+public:
+    int removeElement(vector<int>& nums, int val) {
+        int i ,count = 0,j = 0,numsSize = nums.size();
+        for(i = 0;i < numsSize;i++)
+        {
+            if(nums[i] == val)
+            { 
+                count++;
+            }
+            else
+                nums[j++] = nums[i];
+        }
+        return numsSize - count;
+    }
+};
+```

docs/Leetcode_Solutions/C++/031_.Next_Permutatio.md

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+# 31.Next Permutatio
+
+**<font color=red>难度Medium</font>**
+
+## 刷题内容
+> 原题连接
+
+* https://leetcode.com/problems/next-permutation/
+
+> 内容描述
+
+```
+Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.
+
+If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).
+
+The replacement must be in-place and use only constant extra memory.
+
+Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
+
+1,2,3 → 1,3,2
+3,2,1 → 1,2,3
+1,1,5 → 1,5,1
+```
+> 思路
+******- 时间复杂度: O(n)******- 空间复杂度: O(1)******
+
+我们可以用两个指针表示需要交换的两个数，遍历数组。这题的最坏的情况下，数组降序排列，排序算法的复杂度也是O(n)。
+
+```cpp
+class Solution {
+public:
+    void nextPermutation(vector<int>& nums) {
+        int n1 = 0,n2 = 0;
+        for(int i = 1;i < nums.size();++i)
+            if(nums[i] > nums[n2])
+            {
+                n1 = n2;
+                n2 = i;
+            }
+            else if((nums[i] < nums[n2] && nums[i] > nums[n1]) || nums[i] == nums[n2])
+                n2 = i;
+            else if(nums[i] <= nums[n1])
+            {
+                int j = i;
+                for(;j < nums.size() - 1;++j)
+                    if(nums[j + 1] > nums[j])
+                    {
+                        n1 = j;
+                        n2 = j + 1;
+                        break;
+                    }
+                i = j + 1;
+            }
+        if(n1 == n2)
+            sort(nums.begin(),nums.end());
+        else
+        {
+            swap(nums[n1],nums[n2]);
+            sort(nums.begin() + n1 + 1,nums.end());
+        }
+    }
+};
+```