LeetCode OJ_题解(python):035-Search Insert Position【Array】【Easy】

题目:

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Here are few examples.


[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0

 

给定一个有序数组和一个目标值target,如果目标值target在数组中,则返回它的位置;

                                                   如果不在,返回它应该有序插入的位置。


题目思路:

    这是一个标准的二分查找:

              如果target在数组中,返回mid;

              如果没有找到,那么和first和last位置的数比较一下就可以得到答案。 

    


代码:

class Solution(object):
    def searchInsert(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        first = 0
        last = len(nums) - 1
        while first < last:                     #  确保数列有序,即在数列中寻找目标值target,直到循环结束
            mid = (first + last + 1) // 2        #  二分查找
            if nums[mid] == target:
                return mid
            if nums[mid] < target:
                first = mid + 1
            else:
                last = mid - 1
        if  target > nums[last]:               #  目标值大于数组最后一个值
            return last + 1
        if target <= nums[first]:              #  目标值小于数组第一个值
            return first
if __name__ == "__main__":
    s = Solution()
    print(s.searchInsert(nums =[1, 3, 5, 6], target = 5))
原文地址:https://www.cnblogs.com/llw1121/p/6698087.html