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58. 打家劫舍.md

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***你是一个专业的小偷,计划偷窃沿街的房屋。每间房内都藏有一定的现金,影响你偷窃的唯一制约因素就是相邻的房屋装有相互连通的防盗系统,如果两间相邻的房屋在同一晚上被小偷闯入,系统会自动报警。给定一个代表每个房屋存放金额的非负整数数组,计算你不触动警报装置的情况下 ,一夜之内能够偷窃到的最高金额。***
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```
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class Solution:
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def rob(self, nums: List[int]) -> int:
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n = len(nums)
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if n<3:
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return max(nums)
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dp = [0]*n
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dp[0] = nums[0]
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dp[1] = max(nums[:2])
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for i in range(2, n):
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dp[i] = max(dp[i-2]+nums[i], dp[i-1])
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return dp[-1]
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```

59. 岛屿数量.md

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***给你一个由 '1'(陆地)和 '0'(水)组成的的二维网格,请你计算网格中岛屿的数量。岛屿总是被水包围,并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。此外,你可以假设该网格的四条边均被水包围。***
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```
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输入:grid = [
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["1","1","0","0","0"],
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["1","1","0","0","0"],
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["0","0","1","0","0"],
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["0","0","0","1","1"]
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]
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输出:3
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```
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```
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class Solution:
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def numIslands(self, grid: List[List[str]]) -> int:
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count = 0
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#构造递归
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def dfs(i, j):
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#递归终止条件
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if not 0<=i<len(grid) or not 0<=j<len(grid[0]) or grid[i][j] == '0':
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return
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grid[i][j] = '0'
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dfs(i+1,j)
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dfs(i-1,j)
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dfs(i,j+1)
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dfs(i,j-1)
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for i in range(len(grid)):
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for j in range(len(grid[0])):
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if grid[i][j] == '1':
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#调用递归
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dfs(i,j)
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count += 1
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#返回要优化的目标
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return count
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```

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