刷题笔记39-动态规划part02篇

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目录

• 动态规划
  • 62.不同路径
  • 63. 不同路径 II

动态规划

● 62.不同路径
● 63. 不同路径 II

62.不同路径

62. 不同路径

法1：动态规划

    int uniquePaths(int m, int n) {
        if (m == 1 || n ==1) return 1;
//        int dp[n];
//        for (int i = 0; i < n; ++i) dp[i] = 1;
        vector<int> dp(n,1);
        for (int j = 1; j < m; ++j) {
            for (int i = 1; i < n; ++i) {
                dp[i] += dp[i - 1];
            }
        }
        return dp[n - 1];
    }

63. 不同路径 II

63. 不同路径 II
法1：动态规划

    //空间优化版本：后续时间给出
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
//        dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
       /* int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();

        vector<vector<int>> dp(m,vector<int>(n,0));
        for (int i = 0; i < m && obstacleGrid[i][0] == 0; ++i) dp[i][0] = 1;
        for (int i = 0; i < n && obstacleGrid[0][i] == 0; ++i) dp[0][i] = 1;
        for (int i = 1; i < m; ++i) {
            for (int j = 1; j < n; ++j) {
                if (obstacleGrid[i][j] == 1) continue;
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];*/

       //空间优化版

       int n = obstacleGrid[0].size();
       int m = obstacleGrid.size();
       if (obstacleGrid[0][0] == 1)return 0;
        vector<int> dp(n);
        for (int j = 0; j < dp.size(); ++j)
            if (obstacleGrid[0][j] == 1)
                dp[j] = 0;
            else if (j == 0)
                dp[j] = 1;
            else
                dp[j] = dp[j-1];

        for (int j = 1; j < m; ++j)
            for (int i = 0; i < dp.size(); ++i) {
                if (obstacleGrid[j][i] == 1)
                    dp[i] = 0;
                else if (i != 0)
                    dp[i] = dp[i] + dp[i - 1];
            }

        return dp.back();
    }