代码随想录算法训练营第31天|455.分发饼干 376. 摆动序列 53. 最大子序和

455.分发饼干 

题目链接：LeetCode - The World's Leading Online Programming Learning Platform

代码随想录

class Solution(object):
    def findContentChildren(self, g, s):
        """
        :type g: List[int]
        :type s: List[int]
        :rtype: int
        """
        # get length of s and g
        m = len(g)
        n = len(s)
        # set pointer for s and g
        i=0
        j=0
        #set up return result
        result=0
        # sort g and s for looping
        g.sort()
        s.sort()
        # lopp if i is samller than length of g and j si smaller than length of s
        while i < m and j < n:
            # serve the the children who has samller appetite first
            if g[i] <= s[j]:
                result += 1
                i+=1
                j+=1
            #if the current cookie does not satify current child, see if a bigger cookie does 
            else:
                j+=1
        return result


            

376. 摆动序列 

代码随想录

解题思路：

class Solution(object):
    def wiggleMaxLength(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        #set up result, pre value,cur value
        #res is one because it is equal or more than one item and we are
        #calculating difference
        pre=0
        cur=0
        res=1
        #skip index 1 since we want difference
        for i in range(1,len(nums)):
            #calculate current difference
            cur=nums[i]-nums[i-1]
            #if pre*cur equal or smaller than 0
            #means it is it satisfy wiggle we can append into res 
            #and replace pre as cur. 
            #else we keep calculate new difference
            if pre*cur<=0 and cur!=0:
                res+=1
                pre=cur
        return res 

53. 最大子序和 

代码随想录

暴力：

class Solution(object):
    def maxSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        res=-float("inf")
        #first layer for set up start point, second layer for finding max
        for i in range(len(nums)):
            sum_=0
            for j in range(i,len(nums)):
                sum_+=nums[j]
                res=max(res,sum_)
        return res

class Solution(object):
    def maxSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        res=-float("inf")
        sum_=0
        #loop for nums
        for i in range(len(nums)):
            #get sum 
            sum_+=nums[i]
            #if current sum bigger than res, assign it to res
            if sum_>res:
                res=sum_
            #if sum_<= 0, we can restart from 0
            if sum_<= 0:
                sum_=0
        return res