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不同路径 II #64

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louzhedong opened this issue Sep 14, 2018 · 0 comments
Open

不同路径 II #64

louzhedong opened this issue Sep 14, 2018 · 0 comments

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@louzhedong
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习题

出处:LeetCode 算法第63题

一个机器人位于一个 m x n 网格的左上角 (起始点在下图中标记为“Start” )。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角(在下图中标记为“Finish”)。

现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径?

img

网格中的障碍物和空位置分别用 10 来表示。

**说明:**mn 的值均不超过 100。

示例 1:

输入:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
输出: 2
解释:
3x3 网格的正中间有一个障碍物。
从左上角到右下角一共有 2 条不同的路径:
1. 向右 -> 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右 -> 向右

思路

动态规划

解答

/**
 * @param {number[][]} obstacleGrid
 * @return {number}
 */
var uniquePathsWithObstacles = function (obstacleGrid) {
  var m = obstacleGrid.length;
  var n = obstacleGrid[0].length;

  var dp = [];
  for (var i = 0; i < m; i  ) {
    dp[i] = [];
  }

  if (obstacleGrid[0][0] == 1) {
    dp[0][0] = 0;
  } else {
    dp[0][0] = 1;
  }
  for (var i = 1; i < m; i  ) {
    dp[i][0] = (obstacleGrid[i][0] == 1) ? 0 : dp[i - 1][0];
  }

  for (var j = 1; j < n; j  ) {
    dp[0][j] = obstacleGrid[0][j] == 1 ? 0 : dp[0][j - 1];
  }

  for (var i = 1; i < m; i  ) {
    for (var j = 1; j < n; j  ) {
      dp[i][j] = obstacleGrid[i][j] == 1 ? 0 : dp[i - 1][j]   dp[i][j - 1];
    }
  }
  return obstacleGrid[m - 1][n - 1] == 1 ? 0 : dp[m - 1][n - 1];
};
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