对每一个nums[i], 寻找 range(i,n) 范围内比nums[i]大的数,并与之交换 以此类推,位置 i 处就是 第i 小的数 两次for循环,时间复杂度为 def selection_sort(nums): n = len(nums) for i in range(n): for j in range(i, n): if nums[i] > nums[j]: nums[i], nums[j] = nums[j]...
while True: try: n=int(input()) nums=[int(i) for i in input().split()] ...
for i in range(n): num = int(input("输入任意整数>>>")) nums.append(num) # 获取最大值和最小值 imax = max(nums) imin = min(nums) # 采用 str.format() 打印结果 print("输入生成的10个数字为{},最大值为{},最小值为{}".format(nums,imax,imin)) 1. 2. 3. 4. 5. 6. 7. 8...
nums = [] for i in range(0, 30): if i % 3 == 0: nums.append(i) print(nums) 用推导式 nums = [i for i in range(30) if i % 3 == 0] 一个更复杂的例子, nums = [1,2,3] n = [[i, i**2] for i in nums] --- n = [[1,1],[2,4],[3,9]] 实际上,推导式的...
nums.sort() nums.reverse() for i in range(1, len(nums)): nums.pop() total += i*sum(nums) 魔法球能量这个就可以了_牛客网_牛客在手,offer不愁
= nums[i-1]: nums[count] = nums[i] count += 1 return count ``` 时间复杂度分析:由于只需遍历一次数组,所以时间复杂度为O(n),其中n为数组的长度。 空间复杂度分析:由于只使用了常数个辅助变量,所以空间复杂度为O(1)。.zip St**刺痛上传11.19 KB文件格式zip...
(方法1)for i in range(1,10): #i为1-9中的值 string = "" for j in range(1,i+1): #为了去掉i和j取值重复,故变为i+1 string +=str(j) + " * "+str(i) + " = " +str(i*j)+"\t" print(string) (方法2)for i in range(1,10): ...
foriinrange(5): forjinrange(5): ifj ==2andi ==0: break else:# only execute when it's no break in the inner loop continue break 3.协助处理异常 nums = [1,3,0,5] fordenominatorinnums: try: 20/denominator exceptZeroDivisionError: ...
idxdict = {} res = 0 for i in range(n): idxdict[nums[i]] = i #字典key是数,value是数的下标 for i in range(n): if nums[i] != i+1: #如果第i位上的数不是i+1 idx = idxdict[i+1] #寻找i+1的下标 if (idx-i)%2 != 0: #说明数i+1下标的奇偶性和i不同,那么i+1无论...
self.permute_helper(nums, 0, ans)returnansdefpermute_helper(self, nums, start, ans):ifstart ==len(nums): ans.append(list(nums))returnforiinrange(start, len(nums)): nums[i], nums[start]=nums[start], nums[i] self.permute_helper(nums, start+1, ans) ...