# Unique Paths II
- tags: [[DP_Matrix](# "根据动态规划解题的四要素，矩阵类动态规划问题通常可用 f[x][y] 表示从起点走到坐标(x,y)的值")]
### Source
- lintcode: [(115) Unique Paths II](http://www.lintcode.com/en/problem/unique-paths-ii/)
~~~
Follow up for "Unique Paths":
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
For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
[
[0,0,0],
[0,1,0],
[0,0,0]
]
The total number of unique paths is 2.
~~~
### 题解
在上题的基础上加了obstacal这么一个限制条件，那么也就意味着凡是遇到障碍点，其路径数马上变为0，需要注意的是初始化环节和上题有较大不同。首先来看看错误的初始化实现。
### C++ initialization error
~~~
class Solution {
public:
/**
* @param obstacleGrid: A list of lists of integers
* @return: An integer
*/
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
if(obstacleGrid.empty() || obstacleGrid[0].empty()) {
return 0;
}
const int M = obstacleGrid.size();
const int N = obstacleGrid[0].size();
vector<vector<int> > ret(M, vector<int>(N, 0));
for (int i = 0; i != M; ++i) {
if (0 == obstacleGrid[i][0]) {
ret[i][0] = 1;
}
}
for (int i = 0; i != N; ++i) {
if (0 == obstacleGrid[0][i]) {
ret[0][i] = 1;
}
}
for (int i = 1; i != M; ++i) {
for (int j = 1; j != N; ++j) {
if (obstacleGrid[i][j]) {
ret[i][j] = 0;
} else {
ret[i][j] = ret[i -1][j] + ret[i][j - 1];
}
}
}
return ret[M - 1][N - 1];
}
};
~~~
### 源码分析
错误之处在于初始化第0行和第0列时，未考虑到若第0行/列有一个坐标出现障碍物，则当前行/列后的元素路径数均为0！
### C++
~~~
class Solution {
public:
/**
* @param obstacleGrid: A list of lists of integers
* @return: An integer
*/
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
if(obstacleGrid.empty() || obstacleGrid[0].empty()) {
return 0;
}
const int M = obstacleGrid.size();
const int N = obstacleGrid[0].size();
vector<vector<int> > ret(M, vector<int>(N, 0));
for (int i = 0; i != M; ++i) {
if (obstacleGrid[i][0]) {
break;
} else {
ret[i][0] = 1;
}
}
for (int i = 0; i != N; ++i) {
if (obstacleGrid[0][i]) {
break;
} else {
ret[0][i] = 1;
}
}
for (int i = 1; i != M; ++i) {
for (int j = 1; j != N; ++j) {
if (obstacleGrid[i][j]) {
ret[i][j] = 0;
} else {
ret[i][j] = ret[i -1][j] + ret[i][j - 1];
}
}
}
return ret[M - 1][N - 1];
}
};
~~~
### 源码分析
1. 异常处理
1. 初始化二维矩阵(全0阵)，尤其注意遇到障碍物时应`break`跳出当前循环
1. 递推路径数
1. 返回`ret[M - 1][N - 1]`
- Preface
- Part I - Basics
- Basics Data Structure
- String
- Linked List
- Binary Tree
- Huffman Compression
- Queue
- Heap
- Stack
- Set
- Map
- Graph
- Basics Sorting
- Bubble Sort
- Selection Sort
- Insertion Sort
- Merge Sort
- Quick Sort
- Heap Sort
- Bucket Sort
- Counting Sort
- Radix Sort
- Basics Algorithm
- Divide and Conquer
- Binary Search
- Math
- Greatest Common Divisor
- Prime
- Knapsack
- Probability
- Shuffle
- Basics Misc
- Bit Manipulation
- Part II - Coding
- String
- strStr
- Two Strings Are Anagrams
- Compare Strings
- Anagrams
- Longest Common Substring
- Rotate String
- Reverse Words in a String
- Valid Palindrome
- Longest Palindromic Substring
- Space Replacement
- Wildcard Matching
- Length of Last Word
- Count and Say
- Integer Array
- Remove Element
- Zero Sum Subarray
- Subarray Sum K
- Subarray Sum Closest
- Recover Rotated Sorted Array
- Product of Array Exclude Itself
- Partition Array
- First Missing Positive
- 2 Sum
- 3 Sum
- 3 Sum Closest
- Remove Duplicates from Sorted Array
- Remove Duplicates from Sorted Array II
- Merge Sorted Array
- Merge Sorted Array II
- Median
- Partition Array by Odd and Even
- Kth Largest Element
- Binary Search
- Binary Search
- Search Insert Position
- Search for a Range
- First Bad Version
- Search a 2D Matrix
- Search a 2D Matrix II
- Find Peak Element
- Search in Rotated Sorted Array
- Search in Rotated Sorted Array II
- Find Minimum in Rotated Sorted Array
- Find Minimum in Rotated Sorted Array II
- Median of two Sorted Arrays
- Sqrt x
- Wood Cut
- Math and Bit Manipulation
- Single Number
- Single Number II
- Single Number III
- O1 Check Power of 2
- Convert Integer A to Integer B
- Factorial Trailing Zeroes
- Unique Binary Search Trees
- Update Bits
- Fast Power
- Hash Function
- Count 1 in Binary
- Fibonacci
- A plus B Problem
- Print Numbers by Recursion
- Majority Number
- Majority Number II
- Majority Number III
- Digit Counts
- Ugly Number
- Plus One
- Linked List
- Remove Duplicates from Sorted List
- Remove Duplicates from Sorted List II
- Remove Duplicates from Unsorted List
- Partition List
- Two Lists Sum
- Two Lists Sum Advanced
- Remove Nth Node From End of List
- Linked List Cycle
- Linked List Cycle II
- Reverse Linked List
- Reverse Linked List II
- Merge Two Sorted Lists
- Merge k Sorted Lists
- Reorder List
- Copy List with Random Pointer
- Sort List
- Insertion Sort List
- Check if a singly linked list is palindrome
- Delete Node in the Middle of Singly Linked List
- Rotate List
- Swap Nodes in Pairs
- Remove Linked List Elements
- Binary Tree
- Binary Tree Preorder Traversal
- Binary Tree Inorder Traversal
- Binary Tree Postorder Traversal
- Binary Tree Level Order Traversal
- Binary Tree Level Order Traversal II
- Maximum Depth of Binary Tree
- Balanced Binary Tree
- Binary Tree Maximum Path Sum
- Lowest Common Ancestor
- Invert Binary Tree
- Diameter of a Binary Tree
- Construct Binary Tree from Preorder and Inorder Traversal
- Construct Binary Tree from Inorder and Postorder Traversal
- Subtree
- Binary Tree Zigzag Level Order Traversal
- Binary Tree Serialization
- Binary Search Tree
- Insert Node in a Binary Search Tree
- Validate Binary Search Tree
- Search Range in Binary Search Tree
- Convert Sorted Array to Binary Search Tree
- Convert Sorted List to Binary Search Tree
- Binary Search Tree Iterator
- Exhaustive Search
- Subsets
- Unique Subsets
- Permutations
- Unique Permutations
- Next Permutation
- Previous Permuation
- Unique Binary Search Trees II
- Permutation Index
- Permutation Index II
- Permutation Sequence
- Palindrome Partitioning
- Combinations
- Combination Sum
- Combination Sum II
- Minimum Depth of Binary Tree
- Word Search
- Dynamic Programming
- Triangle
- Backpack
- Backpack II
- Minimum Path Sum
- Unique Paths
- Unique Paths II
- Climbing Stairs
- Jump Game
- Word Break
- Longest Increasing Subsequence
- Palindrome Partitioning II
- Longest Common Subsequence
- Edit Distance
- Jump Game II
- Best Time to Buy and Sell Stock
- Best Time to Buy and Sell Stock II
- Best Time to Buy and Sell Stock III
- Best Time to Buy and Sell Stock IV
- Distinct Subsequences
- Interleaving String
- Maximum Subarray
- Maximum Subarray II
- Longest Increasing Continuous subsequence
- Longest Increasing Continuous subsequence II
- Graph
- Find the Connected Component in the Undirected Graph
- Route Between Two Nodes in Graph
- Topological Sorting
- Word Ladder
- Bipartial Graph Part I
- Data Structure
- Implement Queue by Two Stacks
- Min Stack
- Sliding Window Maximum
- Longest Words
- Heapify
- Problem Misc
- Nuts and Bolts Problem
- String to Integer
- Insert Interval
- Merge Intervals
- Minimum Subarray
- Matrix Zigzag Traversal
- Valid Sudoku
- Add Binary
- Reverse Integer
- Gray Code
- Find the Missing Number
- Minimum Window Substring
- Continuous Subarray Sum
- Continuous Subarray Sum II
- Longest Consecutive Sequence
- Part III - Contest
- Google APAC
- APAC 2015 Round B
- Problem A. Password Attacker
- Microsoft
- Microsoft 2015 April
- Problem A. Magic Box
- Problem B. Professor Q's Software
- Problem C. Islands Travel
- Problem D. Recruitment
- Microsoft 2015 April 2
- Problem A. Lucky Substrings
- Problem B. Numeric Keypad
- Problem C. Spring Outing
- Microsoft 2015 September 2
- Problem A. Farthest Point
- Appendix I Interview and Resume
- Interview
- Resume