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CSメモ
身につけた良さそうなアルゴリズム
- DFS
- BFS
- 平衡二分木
- ソート
- 2分探索
- 動的計画法
- ビット演算
１秒間で処理できるfor文の回数=10^8
バケット法
num[i]=値iが何個あるか
柴確認
Codility
BinaryGap
二進数にした時、0が何回連続するか答えよ
def solution(N):
# write your code in Python 3.6
b=bin(N)[2:]
cnt=0
max_cnt=0
zero_cnt=0
for i in b:
if i=='0':
cnt+=1
if max_cnt<cnt:
max_cnt=cnt
if i=='1':
cnt=0
zero_cnt+=1
if zero_cnt<2:
max_cnt=0
return max_cnt
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