題目描述:
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.
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
.
Note: m and n will be at most 100.
分析:
題意:給定一個m x n 二維地圖(元素0代表空格、元素1代表障礙),左上角爲起點,右下角爲終點,一個機器人只能往下、往右走。返回到達終點所有不重複的路徑條數。
思路:這道題是LeetCode 62的進化版本,我們採用動態規劃的方法:我們構造數組dp[m][n],dp[i][j]表示從位置(0, 0)到達位置(i, j)的路徑條數。初始化各元素值爲0(其中dp[0][0] = 1,dp[i][0] = 1(i∈1→m - 1,出現該列元素爲1之前),dp[0][j] = 1(j∈1→n - 1,出現該行元素爲1之前)),那麼對於位置(i, j),根據題意,Ⅰ. 如果地圖對應值爲0,那麼只能從其上方或者左側移動過來兩種可能;Ⅱ. 如果地圖對應值爲1,說明是障礙,方式爲0種。因此狀態轉移方程爲:
dp[i][j] = (obstacleGrid[i][j] == 0)? dp[i - 1][j] + dp[i][j - 1]: 0。
時間複雜度爲O(m * n)。
代碼:
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
// Exceptional Case:
if(obstacleGrid.empty()){
return 0;
}
int m = obstacleGrid.size(), n = obstacleGrid[0].size();
if(obstacleGrid[0][0] == 1 || obstacleGrid[m - 1][n - 1] == 1){
return 0;
}
// create
vector<vector<int>> dp(m, vector<int>(n, 0));
// init
dp[0][0] = 1;
int pos = 1;
while(pos <= n - 1 && obstacleGrid[0][pos] == 0){
dp[0][pos] = 1;
pos++;
}
pos = 1;
while(pos <= m - 1 && obstacleGrid[pos][0] == 0){
dp[pos][0] = 1;
pos++;
}
// dp
for(int i = 1; i <= m - 1; i++){
for(int j = 1; j <= n - 1; j++){
dp[i][j] = (obstacleGrid[i][j] == 0)? dp[i - 1][j] + dp[i][j - 1]: 0;
}
}
// get answer
return dp[m - 1][n - 1];
}
};