for i in range(1,n): while i>0 and nums[i]<nums[i-1]: nums[i-1],nums[i]=nums[i],nums[i-1] i-=1 return nums 1. 2. 3. 4. 5. 6. 7. 8. 或者 class Solution(object): def sortArray(self, nums): n=len(nums) for i in range(1,n): for j in range(i-1,-1,-1...
nums=[2,3,5,1,9] for i in range(n): print(nums[i]) 1. 2. 3. 4. 2 3 5 1 9 1. 2. 3. 4. 5. 所以直接放心用就好了。range内参数是数组长度,使用i遍历一遍,就会直接全部输出 nums=[2,3,5,1,9] for i in range(len(nums)): print(nums[i]) 1. 2. 3. 2 3 5 1 9 1....
有数组: nums = [1, 2, 3, 4] 如果要计算每个元素数值在数组总和的百分比,正确的代码写法有: A、nums = [1, 2, 3, 4] numsNew = [0] * 4 for i in range(len(nums)): numsNew[i] = nums[i] / sum(nums) print(numsNew) B、nums = [1, 2, 3, 4] for i in range(len(nu
if 2*k>=len(prices): return sum([max(0, prices[i]-prices[i-1]) for i in range(1, len(prices))]) # val在每一次循环中,代表at most k1次交易,且右边卖出点在i的最大收益 # 于是 pnl[i] 也就取val 和 pnl[i-1] val = 0 pnl = [0] * len(prices) for _ in range(k): val ...
(方法1)for i in range(1, 9): 循环 for v in range(1, 9): 嵌套循环 if i !=v: 去重 count += 1 # count = count -8(也为去重作用,因有8个元素,故共8个重复的) print(count) (方法2)count =0 ls = [3,5,8,9] for i in range(0, len(ls)): ...
] if len(nums)==0: print(0) for i in nums: position = bisect.bisect_lef...
for i in range(4, 0, -1): print(" "*(5-i) + "* "*(i)) 2.使用for循环实现冒泡排序。 答案: nums = [5, 2, 7, 1, 9] for i in range(len(nums)-1): for j in range(len(nums)-1-i): if nums[j] > nums[j+1]: ...
题目code-"0123456789ABCDEFnums[r=inpui ("请输入一个二进制整数:")numlen= len(numstr)s-0ss=s+1^2/ fori in range(numlen):x=int(numstr(i])*2**(numlen-i-1)S+-xI-swhile t0:ss=code[t%16]+stt 16print(s) 相关知识点: 试题来源: ...
forkey, iteminuser.items(): print(key, item) range range 的作用是生成一个类似于列表的数据,range(6) 生成类似于 [0,1,2,3,4,5] 的数据。当你需要对某段代码循环运行指定次数,但是又没有现成的数据可以遍历时,可以用 range foritem inrange(10000):print(item) ...
流程图: 实例:记录及格人数 scores=[100,30,60,70,33,60,55,80]nums=0for i in scores:ifi<60:continueelse: nums+=1print('及格人数:%d个人'%nums) 四、参考资料 《python王者归来》 洪锦魁著