[leetcode] 63. Unique Paths II (medium)

原题
思路：
用到dp的思想，到row,col点路径数量 ：

path[row][col]=path[row][col-1]+path[row-1][col];

遍历row*col，如果map[row][col]为1，则将其置为0；如果非1，则进行上述公式。
最后返回path[终点row][终点col]的值即为解。

一开始的代码，beat 44%，效率不高

class Solution {
 public:
  int uniquePathsWithObstacles(vector<vector<int>> &obstacleGrid) {
    int rowSize = obstacleGrid.size();
    int colSize = obstacleGrid[0].size();

    if (obstacleGrid[0][0] == 1) return 0;

    obstacleGrid[0][0] = 1;
    for (int col = 1; col < colSize; col++) {
      if (obstacleGrid[0][col] == 1)
        obstacleGrid[0][col] = 0;
      else
        obstacleGrid[0][col] += obstacleGrid[0][col - 1];
    }
    for (int row = 1; row < rowSize; row++) {
      if (obstacleGrid[row][0] == 1)
        obstacleGrid[row][0] = 0;
      else
        obstacleGrid[row][0] += obstacleGrid[row-1][0];
    }
    for (int i = 1; i < rowSize; i++) {
      for (int j = 1; j < colSize; j++) {
        if (obstacleGrid[i][j] == 1)
          obstacleGrid[i][j] = 0;
        else
          obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
      }
    }
    return obstacleGrid[rowSize - 1][colSize - 1];
  }
};

优化后的代码：
使用额外一个数组记录走到每一列有几种走法，因为题目只求终点，则使用一维数组即可。
使用一个额外的整型pre记录当前的前一步有多少种走法。（左边走来+上边走来）
则有：

pre(当前)=dp[col](上一行当前列) + pre(左一格当前行);
dp[col](当前)=pre;

beat 100%

class Solution {
 public:
  int uniquePathsWithObstacles(vector<vector<int>> &obstacleGrid) {
    int rowSize = obstacleGrid.size();
    int colSize = obstacleGrid[0].size();
    int dp[colSize] = {0};
    int pre = 1;
    for (int i = 0; i < rowSize; i++) {
      for (int j = 0; j < colSize; j++) {
        if (obstacleGrid[i][j] == 0) {
          pre += dp[j];
          dp[j] = pre;
        }
        else {
          pre = 0;
          dp[j] = 0;
        }
      }
      pre=0;
    }
    return dp[colSize-1];
  }
};