Leetcode-动态规划

八、动态规划

1. 斐波那契数列

方法一：Fib(n) = Fib(n - 1) + Fib(n - 2), 可采用递归的方法，获取n的值

方法二：定义两个变量mm和nn用来表示n-1的值和n-2的值，只需要O(n)的时间复杂度和O(1)的空间复杂度即可得到n的值

		if(n == 0 || n == 1)
	    {
   
	        return n;
	    }
	
	    int a = 1; int b = 0;
	    for(int i = 2; i <= n; i++)
	    {
   
	        int temp = a;
	        a = (a + b)% 1000000007;
	        b = temp;
	    }
	
	    return a;

2. 青蛙跳台阶问题

本质上就是斐波那契数列

		if(n == 0 || n == 1){
   
	        return 1;
	    }
	    int res = 0;
	    int mm = 1;
	    int nn = 1;
	    for(int i = 2; i <= n; i++){
   
	        res = (mm + nn ) % 1000000007;
	        nn = mm;
	        mm = res;
	    }
	
	    return res;

3. 最小路径和

创建一个一维数组，用来记录对当前列来说的最短路径和，按行遍历，获取当前列的最短路径和。

		if(grid == null || grid[0] == null || grid.Length == 0 || grid[0].Length == 0){
   
            return 0;
        }

        int[] dp = new int[grid[0].Length];
        dp[0] = grid[0][0];
        for(int i = 1; i < grid[0].Length; i++){
   
            dp[i] = dp[i - 1] + grid[0][i];
        }
        int left = int.MaxValue;
        for(int i = 1;i < grid.Length; i++){
   
            left = grid[i][0] + dp[0];
            dp[0] = left;
            for(int j = 1; j < grid[i].Length; j++){
   
                dp[j] = dp[j] < left ? dp[j] + grid[i][j] : left + grid[i][j];
                left = dp[j];
            }

        }
        return dp[grid[0]