网易互娱笔试0ac代码
# 插入子串成为回文串 91.3% class Solution: def canBePalindromicString(self, str1): n = len(str1) left = 0 right = n - 1 while left < right: if str1[left] == str1[right]: n -= 2 left += 1 right -= 1 if n <= 5: return 1 else: return 0 # 最长连续1的长度 90% C, K = map(int, input().split()) c = str(C) n = len(c) ANS = [] for i in range(K, n): K1 = K ans = 0 for j in range(i, -1, -1): if c[j] == '1': ans += 1 else: K1 -= 1 if K1 >= 0: ans += 1 if K1 < 0: break ANS.append(ans) print(max(ANS)) # 航班选座系统 10% N, C = map(int, input().split()) zuowei = [] for i in range(N): zuowei.append(list(map(str, input().split()))) mat = [[0]*7 for _ in range(N)] ans = 0 # for i in range(N): # for j in range(9): # if zuowei[i][0][j] == '.' and j < 3: # mat[i][j] = 1 # ans += 1 # if zuowei[i][0][j] == '.' and j > 5: # mat[i][j-2] = 1 # ans += 1 # if ans > C: # print('SUCCESS') # for i in range(N): # for j in range(9): # if zuowei[i][0][j] == '.' and j == 0: # print(str(j+1)+'A') # C -= 1 # if zuowei[i][0][j] == '.' and j == 1: # print(str(j+1)+'B') # C -= 1 # if zuowei[i][0][j] == '.' and j == 2: # print(str(j+1)+'C') # C -= 1 # if zuowei[i][0][j] == '.' and j == 6: # print(str(j-2)+'D') # C -= 1 # if zuowei[i][0][j] == '.' and j == 7: # print(str(j-2)+'E') # C -= 1 # if zuowei[i][0][j] == '.' and j == 8: # print(str(j-2)+'F') # C -= 1 # if C == 0: # break # if C == 0: # break # else: print('FAILED') # 升序数组第k小 66.67% class Solution: def find_kth(self, arr1, arr2, k): arr = arr1 + arr2 arr.sort() return arr[k-1]
#网易互娱##笔经#