n=len(nums) 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...
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
def first_missing_positive(nums): for i in range(len(nums)): while 1 <= nums[i] <= len(nums) and nums[i] != nums[nums[i] - 1]: idx = nums[i] - 1 nums[i], nums[idx] = nums[idx], nums[i] for i in range(len(nums)): if nums[i] != i + 1: return i + 1 re...
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]: ...
for i in range(1,11,2): 从1开始,每次加2,到9结束 for i in range(1,1): 这个循环一次都不执行 while True: 构建死循环用while nums = [0,1,2,3,4,5,6,7,8,9] for循环的对象也可为一个集合(列表,字典等) for num in nums:
for i in range(len(nums)): lo, hi = i + 1, len(nums) - 1 while lo < hi: sum = nums[i] + nums[lo] + nums[hi] if abs(target - sum) < abs(diff): diff = target - sum if sum < target: lo += 1 else: hi -= 1 ...
while True: try: n=int(input()) nums=[int(i) for i in input().split()] ...
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
for i in range(5): print(i) for循环的等价形式是while循环: python nums = [1, 2, 3, 4, 5] for num in nums: print(num) 等价的while循环形式: python nums = [1, 2, 3, 4, 5] i = 0 while i < len(nums): print(nums[i]) i += 1 无论是while循环还是for循环,它们都可以用来...
for i in range(len(right_max)): res = res + max(min(right_max[i],left_max[i])-height[i],0) return res Stack Method Algorithm 这个算法我非常非常不喜欢,因为很不好理解,非常不推荐 属于每一步都要判断能否有横切的机会 class Solution: ...