Leetcode#63||Unique Paths II
2015-08-18 11:23
316 查看
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length ==0) {
return 0;
}
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[] paths = new int
;
for (int j = 0; j < n; j++) {
if (obstacleGrid[0][j] == 0) {
paths[j] = 1;
} else if (obstacleGrid[0][j] == 1) {
break;
}
}
for (int i = 1; i < m; i++) {
if (obstacleGrid[i][0] == 1) {
paths[0] = 0;
}
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] == 1) {
paths[j] = 0;
} else {
paths[j] += paths[j - 1];
}
}
}
return paths[n - 1];
}
}
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid == null || obstacleGrid.length == 0 || obstacleGrid[0].length ==0) {
return 0;
}
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[] paths = new int
;
for (int j = 0; j < n; j++) {
if (obstacleGrid[0][j] == 0) {
paths[j] = 1;
} else if (obstacleGrid[0][j] == 1) {
break;
}
}
for (int i = 1; i < m; i++) {
if (obstacleGrid[i][0] == 1) {
paths[0] = 0;
}
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] == 1) {
paths[j] = 0;
} else {
paths[j] += paths[j - 1];
}
}
}
return paths[n - 1];
}
}
相关文章推荐
- 解析从源码分析常见的基于Array的数据结构动态扩容机制的详解
- javascript数组操作总结和属性、方法介绍
- mysql_fetch_assoc和mysql_fetch_row的功能加起来就是mysql_fetch_array
- JavaScript Array扩展实现代码
- JavaScript之数组(Array)详解
- C#中Array与ArrayList用法及转换的方法
- Array栈方法和队列方法的特点说明
- Array.prototype.slice 使用扩展
- Array, Array Constructor, for in loop, typeof, instanceOf
- 实例详解ECMAScript5中新增的Array方法
- js Array的用法总结
- JavaScript 判断判断某个对象是Object还是一个Array
- Javascript中的Array数组对象详谈
- js模拟实现Array的sort方法
- 在javascript将NodeList作为Array数组处理的方法
- PHP array_multisort()函数的使用札记
- PHP中array_merge和array相加的区别分析
- Sorting Array Values in PHP(数组排序)
- 深入array multisort排序原理的详解
- Array 重排序方法和操作方法的简单实例