def sortArray(self, nums: List[int]): for i in range(len(nums)-1): minindex=i for j in range(i+1,len(nums)): if nums[j]<nums[minindex]: minindex=j nums[i],nums[minindex]=nums[minindex],nums[i] return nums 1. 2. 3. 4. 5. 6. 7. 8. 冒泡排序 比较相邻元素的大小,...
def searchInsert(self, nums: List[int], target: int) -> int: left, right = 0, len(nums) - 1 while left <= right: middle = (left + right) // 2 if nums[middle] < target: left = middle + 1 elif nums[middle] > target: right = middle - 1 else: return middle return right ...
因此,有两种解决办法:1是倒序遍历range(len(nums)-1,-1,-1) 从后面开始删,list在变短,我们访问的下标也在变小。2.遍历一个不变的list,即原来目标的拷贝nums[:]。这样我们就不存在越界的 智能推荐 成功解决sys.argv[1] IndexError: list index out of range错误...
Python Programming, simpleSearching Problem simplesearching function: def search(x, nums): listwhere somesample interactions: -1Python Programming, SimpleSearching Problem firstexample, functionreturns indexwhere secondexample, returnvalue -1 indicates Pythonincludes built-insearch-related methods! Python ...
>>> print(1, 2, 3) 1 2 3 >>> 我们也可以按需修改。 | >>> print(1, 2, 3, sep='|') 1|2|3 >>> \n >>> print(1, 2, 3, sep='\n') 1 2 3 >>> \n{'-' * 10}\n >>> print(1, 2, 3, sep=f"\n{'-' * 10}\n") 1 --- 2 --- 3 >>> 值得留意,我们控...
index # 开始位置与结束位置 while end <= len(nums) - 1: if nums[end] == target: # 向右查找 end += 1 else: break while start >= 0: if nums[start] == target: # 向左查找 start -= 1 else: break return [start + 1, end - 1] def binary_search(self, low, high, nums, targ...
Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.Your algorithm's runtime complexity must be in the order of O(log n).If the target is not found in the array, return [-1, -1]....
这个是不是因为for循环range()函数中只会在第一次调用时计算len(nums),之后range的值就不会改变了。 相关代码 // 请把代码文本粘贴到下方(请勿用图片代替代码)这是for循环的: 1 nums = [1,1,1,2,3,1,1] 2 count = 0 3 if len(nums) == 1 or len(nums) == 0: 4 count = len(nums) 5 ...
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
classSolution(object):defsearchRange(self, nums, target):""":type nums: List[int] :type target: int :rtype: List[int]"""ifnotnums:return[-1,-1] l=0 r= len(nums)-1whilel+1 < r:#find leftmid = l + (r-l)/2ifnums[mid] ==target: ...