My solution from the contest:
def findContentChildren(self, g, s): g.sort() s.sort() res = 0 i = 0 for e in s: if i == len(g): break if e >= g[i]: res += 1 i += 1 return res
O(nlogn) time and O(1) space
@dalwise excuse me.why use "res" and "i"?
It is not O(NlogN). It's O(NlogN+MlogM)
def findContentChildren(self, g, s): i, j, g, s = 0, 0, sorted(g), sorted(s) while i < len(g) and j < len(s): i += g[i] <= s[j] j += 1 return i
@dalwise yes,your solution is great,and my thoughts is same,this is my code:
class Solution(object): def findContentChildren(self, g, s): g=sorted(g);s=sorted(s) count=0;i=0 while count<len(g) and i<len(s): if g[count]<=s[i]:count+=1 i+=1 return count
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