一天两道力扣（6）

class Solution(object):
    def numIslands(self, grid):
        def dfs(grid, i, j):
            if not 0 <= i < len(grid) or not 0 <= j < len(grid[0]) or grid[i][j] == '0': return
            grid[i][j] = '0'
            dfs(grid, i + 1, j)
            dfs(grid, i - 1, j)
            dfs(grid, i, j + 1)
            dfs(grid, i, j - 1)
        cnt = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == '1':
                    dfs(grid, i, j)
                    cnt += 1
        return cnt
        

思路：就是从头开始遍历，每次遇到一个值为1的，就给它上下左右及自身均为1的变为0（这里用DFS深度优先遍历），然后cnt加一，这就算一个岛屿，直到遍历结束

class Solution(object):
    def rob(self, nums):
        if len(nums) == 0:
            return 0
        N = len(nums)
        dp = [0] * (N+1)
        dp[1] = nums[0]
        dp[0] = 0
        for i in range(2, N+1):
            dp[i] = max(dp[i - 1], nums[i - 1] + dp[i - 2])
        return dp[N]

思路：

转移方程：（1）偷前i-1间房子，最后一间不偷（2）偷前i-2间房子和最后一间