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. Notice m and n will be at most 100. Have you met this question in a real interview? Yes 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. Tags
初始化, if 判断不同的状态
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
// write your code here
if (obstacleGrid == null || obstacleGrid.length == 0) {
return -1;
}
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
//state
int[][] sum = new int[m][n];
//initialize
if (obstacleGrid[0][0] == 1) {
sum[0][0] = 0;
} else {
sum[0][0] = 1;
}
for (int i = 1; i < n; i++) {
if (obstacleGrid[0][i] == 1 || sum[0][i - 1] == 0) {
sum[0][i] = 0;
} else {
sum[0][i] = 1;
}
}
for (int i = 1; i < m; i++) {
if (obstacleGrid[i][0] == 1 || sum[i - 1][0] == 0) {
sum[i][0] = 0;
} else {
sum[i][0] = 1;
}
}
//function
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] == 1) {
sum[i][j] = 0;
} else {
sum[i][j] = sum[i - 1][j] + sum[i][j - 1];
}
}
}
//ans
return sum[m - 1][n - 1];
}
原文：http://www.cnblogs.com/apanda009/p/7290888.html