Merge pull request #252 from xshahq/master

第63题和66题

docs/Leetcode_Solutions/C++/0063._Unique_Paths_II.md

+# 63. Unique Paths II
+
+**<font color=red>难度:Medium<font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/unique-paths-ii/
+
+> 内容描述
+
+```
+A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
+
+The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
+
+Now consider if some obstacles are added to the grids. How many unique paths would there be?
+
+
+
+An obstacle and empty space is marked as 1 and 0 respectively in the grid.
+
+Note: m and n will be at most 100.
+
+Example 1:
+
+Input:
+[
+  [0,0,0],
+  [0,1,0],
+  [0,0,0]
+]
+Output: 2
+Explanation:
+There is one obstacle in the middle of the 3x3 grid above.
+There are two ways to reach the bottom-right corner:
+1. Right -> Right -> Down -> Down
+2. Down -> Down -> Right -> Right
+
+```
+
+
+> 思路1
+
+******- 时间复杂度: O(n^2)******- 空间复杂度: O(n^2)******
+
+运用动态规划的思想，写出状态转移方程，d(i,j) = d(i + 1,j) + d(i,j + 1)，不过要注意边界值
+
+```cpp
+class Solution {
+public:
+    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
+        int length1 = obstacleGrid.size(),length2 = obstacleGrid[0].size();
+        long long dp[length1][length2];
+        memset(dp,0,sizeof(dp));
+        for(int i = length1 - 1;i >= 0;--i)
+            for(int j = length2 - 1;j >= 0;--j)
+            {
+                if(!obstacleGrid[i][j])
+                {
+                    if(i == length1 - 1 && j == length2 - 1)
+                    dp[i][j] = 1;
+                else
+                {
+                    if(i == length1 - 1)
+                        dp[i][j] = dp[i][j + 1];
+                    else if(j == length2 - 1)
+                        dp[i][j] = dp[i + 1][j];
+                    else
+                        dp[i][j] = dp[i][j + 1] + dp[i + 1][j];
+                }
+            }
+        }
+    return dp[0][0];
+    }
+};
+```

docs/Leetcode_Solutions/C++/0066._Plus_One.md

+# 66. Plus One
+
+**<font color=red>难度:Easy<font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/plus-one/
+
+> 内容描述
+
+```
+Given a non-empty array of digits representing a non-negative integer, plus one to the integer.
+
+The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.
+
+You may assume the integer does not contain any leading zero, except the number 0 itself.
+
+Example 1:
+
+Input: [1,2,3]
+Output: [1,2,4]
+Explanation: The array represents the integer 123.
+Example 2:
+
+Input: [4,3,2,1]
+Output: [4,3,2,2]
+Explanation: The array represents the integer 4321.
+
+```
+
+
+> 思路1
+
+******- 时间复杂度: O(n)******- 空间复杂度: O(1)******
+
+将字符转成数字即可，不过这里要注意有可能会进一
+
+```cpp
+class Solution {
+public:
+    vector<int> plusOne(vector<int>& digits) {
+        int i = digits.size() - 1;
+        while((i >= 0) &&  (digits[i] == 9))
+        {
+            digits[i] = 0;
+            --i;
+        }
+        if(i < 0)
+            digits.insert(digits.begin(),1);
+        else
+            ++digits[i];
+        return digits;
+    }
+};
+```
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