以下为暴力查找,更优的方法应该用二分法查找 class Solution(object): def searchInsert(self, nums, target): if not nums: return 0 if nums[0]>target: return 0 if nums[len(nums)-1]<target: return len(nums) for i in range(len(nums)-1): if nums[i]==target: return i if nums[i] < target and nums[i+1] >target: return i+1 nums=[1,3,5,6] target=5 object=Solution() object1=object.searchInsert(nums,target) print(object1) nums=[1,3,5,6] target=2 object=Solution() object1=object.searchInsert(nums,target) print(object1) nums=[1,3,5,6] target=7 object=Solution() object1=object.searchInsert(nums,target) print(object1) nums=[1,3,5,6] target=0 object=Solution() object1=object.searchInsert(nums,target) print(object1) nums=[1,3,5,6] target=1 object=Solution() object1=object.searchInsert(nums,target) print(object1) 运行: 2 1 4 0 0 Process finished with exit code 0 ''' 给定一个排序数组和一个目标值,在数组中找到目标值,并返回其索引。如果目标值不存在于数组中,返回它将会被按顺序插入的位置。 你可以假设数组中无重复元素。 示例 1: 输入: [1,3,5,6], 5 输出: 2 示例 2: 输入: [1,3,5,6], 2 输出: 1 示例 3: 输入: [1,3,5,6], 7 输出: 4 示例 4: 输入: [1,3,5,6], 0 输出: 0 '''