【经典LeetCode算法题目专栏分类】【第5期】贪心算法:分发饼干、跳跃游戏、模拟行走机器人

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分发饼干

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| class Solution : def findContentChildren****(**** self****,**** g****:**** List****[**** int ], s****:**** List****[**** int ]) -> int : # 贪心算法 res = 0 g****.**** sort****()**** s****.**** sort****()**** i = 0 j = 0 while i < len ( g****)**** and j < len ( s****):**** # 饼干满足胃口 if g****[**** i****]**** <= s****[**** j****]:**** res += 1 i += 1 j += 1 else : # 饼干不满足胃口,查找下一个饼干 j += 1 return res |

跳跃游戏

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| class Solution : def canJump****(**** self****,**** nums****:**** List****[**** int ]) -> bool : # 贪心算法 reach_index = len ( nums****)**** - 1 # 表示能够到达的索引位置 for i in range ( len ( nums****)-**** 1****,-**** 1****,-**** 1****):**** # 从后往前遍历,如果满足下述条件说明能够达到当前索引 if i + nums****[**** i****]**** >= reach_index****:**** reach_index = i return reach_index == 0 class Solution : def canJump****(**** self****,**** nums****:**** List****[**** int ]) -> bool : if nums == [ 0****]:**** return True maxDist = 0 # 能够达到的最远距离 end_index = len ( nums****)-**** 1 for i****,**** jump in enumerate ( nums****):**** # maxDist >= i表示能够达到当前索引位置,并且从当前索引开始 if maxDist >= i and i****+**** jump > maxDist****:**** maxDist = i****+**** jump if maxDist >= end_index****:**** return True return False |

跳跃游戏2

|--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| class Solution : def jump****(**** self****,**** nums****:**** List****[**** int ]) -> int : end = 0 # end 表示当前能跳的边界 maxPosition = 0 steps = 0 for i in range ( len ( nums****)**** - 1****):**** # 找能跳的最远的 maxPosition = max ( maxPosition****,**** nums****[**** i****]**** + i****);**** if i == end****:**** #遇到边界,就更新边界,并且步数加一 end = maxPosition****;**** steps += 1 return steps****;**** |

模拟行走机器人

|---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| class Solution : def robotSim****(**** self****,**** commands****:**** List****[**** int ], obstacles****:**** List****[**** List****[**** int ]]) -> int : if not commands****:**** return 0 # 索引0,1,2,3分别表示北,东,南,西 direx = [ 0****,**** 1****,**** 0****,**** - 1****]**** direy = [ 1****,**** 0****,**** - 1****,**** 0****]**** curx****,**** cury****,**** curdire****,**** ans = 0****,**** 0****,**** 0****,**** 0 com_len****,**** obs_len = len ( commands****),**** len ( obstacles****)**** obstacle_set = {( obstacles****[**** i****][**** 0****],**** obstacles****[**** i****][**** 1****])**** for i in range ( obs_len****)}**** # 变为集合,使判断是否有障碍物更快 for i in range ( com_len****):**** if commands****[**** i****]**** == - 1****:**** # 向右转90度 curdire = ( curdire + 1****)**** % 4 elif commands****[**** i****]**** == - 2****:**** # 向左转90度 curdire = ( curdire + 3****)**** % 4 else : # 1 <= x <= 9: 向前移动x个单位长度 for j in range ( commands****[**** i****]):**** # 试图走出一步,并判断是否遇到了障碍物 nx = curx + direx****[**** curdire****]**** ny = cury + direy****[**** curdire****]**** # 当前坐标不是障碍物,计算并存储的最大欧式距离的平方做比较 if ( nx****,**** ny****)**** not in obstacle_set****:**** curx = nx cury = ny ans = max ( ans****,**** curx********* curx + cury********* cury****)**** else : # 是障碍点,被挡住了,停留,智能等待下一个指令,那可以跳出当前指令了。 break return ans |

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