虾皮4.6算法岗笔试 编程题目大意及AC代码
4.9更新：
依然没有收到面试邀请，可能是简历不匹配，也可能是没hc了。
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之前看大家分享就很焦虑，各种全a也没有面试机会，求求了，给个面试机会吧虾皮。。。
1. 对数据（x，y）有两种操作：两边各加1和两边各乘2，问从（x，y）到（a，b）最少需要几次操作？
bfs
import collections class Solution: def GetMinCalculateCount(self, sourceX, sourceY, targetX, targetY) : q = collections.deque([(targetX, targetY)]) res = 0 visited = set() while q: for _ in range(len(q)): a, b = q.popleft() if a == sourceX and b == sourceY: return res elif a > sourceX and b > sourceY: if (a - 1, b - 1) not in visited: visited.add((a - 1, b - 1)) q.append((a - 1, b - 1)) if a % 2 == 0 and b % 2 == 0 and (a // 2, b // 2) not in visited: visited.add((a // 2, b // 2)) q.append((a // 2, b // 2)) res += 1 return -1 a = Solution() a.GetMinCalculateCount(10,100,22,203)2. 把只含有A和B原字符串变成互逆字符串：
res1 = 0 res2 = 0 s1 = list(s) s2 = list(s) # begin with "A" pre = "B" for i in range(len(s)): if s1[i] == pre: res1 += 1 pre = "A" if pre == "B" else "B" # begin with "B" pre = "A" for i in range(len(s)): if s1[i] == pre: res2 += 1 pre = "A" if pre == "B" else "B" print(min(res1, res2))3. 8皇后原题
n = 4 res = [0] s1 = set() #row s2 = set() #column s3 = set() #diag def backtracking(idx): if idx == n: res[0] += 1 return for i in range(n): if i not in s1 and i + idx not in s2 and i - idx not in s3: s1.add(i) s2.add(i + idx) s3.add(i - idx) backtracking(idx + 1) s1.remove(i) s2.remove(i + idx) s3.remove(i - idx) backtracking(0) print(res[0])
