Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. You may assume no duplicates in the array. Here are few examples. [1,3,5,6], 5 → 2 [1,3,5,6], 2
最优时间复杂度(不可靠) 最坏时间复杂度(保证) 平均时间复杂度(平均状况) 不同语句的时间复杂度: (1)顺序语句:使用加法 (2)循环语句:使用乘法 (3)分支语句:使用坏时间复杂度 例如:如下代码的时间复杂度: #!/usr/bin/env python #! _*_ coding:UTF-8 _*_ for a in range(1001): for b in range(1001): c = 1000 - a - b if a**2 + b**2 == c**2: print "a=%
#!/usr/bin/env python #! _*_ coding:UTF-8 _*_ from Queue import Queue import time que = Queue() time_begin = time.time() # 如果a+b+c=1000, 且a^2+b^2=c^2,a,b,c为自然数,求出a,b,c所有的组合 # 使用枚举法计算结果 for a in range(1001): for b in range(1001): for c in range(1001):
Write a function: def solution(A) that, given an array A of N integers, returns the smallest positive integer (greater than 0) that does not occur in A. For example, given A = [1, 3, 6, 4, 1, 2], the function should return 5. Given A = [1, 2, 3], the
目录 1.算法基础 2.冒泡排序 3.时间复杂度 (1)时间频度 (2)时间复杂度 4.指数时间 5.常数时间 6.对数时间 7.线性时间 1.算法基础 要求:生成一个4*4的2维数组并将其顺时针旋转90度 #!_*_coding:utf-8_*_ array=[[col for col in range(5)] for row in range(5)] #初始化一个4*4数组 #array=[[col for col in 'abcde'] for row in range(5)] for