【每日一题0714】搜索插入位置

给你一个有序的数字列表和一个目标数值,任务是在列表中找到目标值,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。

备注:列表是无重复的,均为正整数。

题目难度:简单
题目来源:leetcode:搜索插入位置

def search_insert(nums:list, target):
    pass

assert search_insert([1,3,5,6], 5) == 2
assert search_insert([1,3,5,6], 2) == 1
assert search_insert([1,3,5,6], 7) == 4
assert search_insert([1,3,5,6], 0) == 0

求简单解法。。

def serach_insert(li_a,n):
    for i in range(len(li_a)):
        if li_a[i]==n:
            return i
        else:
            li_a.append(n)
            li_a.sort()
            for j in range(len(li_a)):
                if li_a[j]==n:
                    return j

assert serach_insert([1,3,5,6], 5) == 2
assert serach_insert([1,3,5,6], 2) == 1
assert serach_insert([1,3,5,6], 7) == 4
assert serach_insert([1,3,5,6], 0) == 0
def search_insert(nums: list, target):
    i = 0
    while i <= len(nums) - 1 and nums[i] < target:
        i += 1
    return i


assert search_insert([1, 3, 5, 6], 5) == 2
assert search_insert([1, 3, 5, 6], 2) == 1
assert search_insert([1, 3, 5, 6], 7) == 4
assert search_insert([1, 3, 5, 6], 0) == 0

def searchInsert(nums, target):
nums.append(target)
nums.sort()
return nums.index(target)



“”"
def search_insert(nums: list, target):
# 优化
print(target)
# 将目标数据插入到列表后面
nums.append(target)
# 将列表排序
nums.sort()
print(nums)
# 返回目标数据索引值(位置)
return nums.index(target)

class TestSearchInsert:
def test_search(self):
assert search_insert([1, 3, 5, 6], 5) == 2
assert search_insert([1, 3, 5, 6], 2) == 1
assert search_insert([1, 3, 5, 6], 0) == 0
assert search_insert([1, 3, 5, 6], 7) == 4

    # or 用循环判断
    # for i in range(len(nums)):
    #
    #     if nums[i] == target:
    #         print(nums)
    #         return i
    #     elif nums[i] > target:
    #         nums.insert(i, target)
    #         print(nums)
    #         return i
    #     elif nums[len(nums)-1]<target:
    #         nums.append(target)
    #         print(nums)
    #         return nums.index(target)

“”"

def search_insert(nums, target):
    """ 搜索插入位置 """
    if target in nums:
        return nums.index(target)
    else:
        nums.append(target)
        nums.sort(reverse=False)
        return nums.index(target)

@15774518534_8534 666666

@fwj 参看这位大佬的优秀思路

@89717045_7695, @13718723532 同学你好,题解都很棒!同时,建议尽量使用markdown文本哈,这样方便其他同学拷贝学习你的代码。具体使用方式是,在回复中,贴源码,然后前、后使用两个``` 符号包围。例如:

my codes here
1 Like
class Solution:
    def search_insert(self: list, target) -> int:
        """
        1、如果目标值在列表中存在,则返回其索引
        2、如果目标值在列表中不存在,返回它将按顺序插入的位置
        :param self:
        :param target:
        :retursn: int:
        """
        if target in self:
            return [i for i, j in list(enumerate(self)) if j == target][0]
        else:
            self.append(target)
            self.sort()
            return self.index(target)

    assert search_insert([1, 3, 5, 6], 5) == 2
    assert search_insert([1, 3, 5, 6], 2) == 1
    assert search_insert([1, 3, 5, 6], 7) == 4
    assert search_insert([1, 3, 5, 6], 0) == 0

:wave: :wave:

@13718723532 酷~~~ :clap: :clap: :clap: :watermelon:

普通青年

def search_insert(nums, target) -> int:
    if target in nums:
        return nums.index(target)
    for i,item in enumerate(nums):
        if target < item:
            return i
        if item < target:
            if i < len(nums) -1:
                continue
            else:
                return i+1

              
assert search_insert([1, 3, 5, 6], 5) == 2
assert search_insert([1, 3, 5, 6], 2) == 1
assert search_insert([1, 3, 5, 6], 7) == 4
assert search_insert([1, 3, 5, 6], 0) == 0

力扣官方题解python版:二分查找

class Solution:
    def search_insert(self, nums, target: int) -> int:
        n = len(nums)
        left, right, ans = 0, n-1, n
    
        while left <= right:
            mid = ((right - left) >> 1) + left
            if target <= nums[mid]:
                ans, right = mid, mid - 1
            else:
                left = mid + 1
        return ans

复杂度分析

时间复杂度:O(logn),其中 n 为数组的长度。二分查找所需的时间复杂度为 O(logn)。
空间复杂度:O(1)。我们只需要常数空间存放若干变量。
参见链接:力扣

:partying_face: :partying_face: :partying_face: :partying_face: :partying_face: :partying_face:

def search_insert(nums: list, target):
    # 优化
    print(target)
    # 将目标数据插入到列表后面
    nums.append(target)
    # 将列表排序
    nums.sort()
    print(nums)
    # 返回目标数据索引值(位置)
    return nums.index(target)

class TestSearchInsert:
    def test_search(self):
        assert search_insert([1, 3, 5, 6], 5) == 2
        assert search_insert([1, 3, 5, 6], 2) == 1
        assert search_insert([1, 3, 5, 6], 0) == 0
        assert search_insert([1, 3, 5, 6], 7) == 4

        # or 用循环判断
        # for i in range(len(nums)):
        #
        #     if nums[i] == target:
        #         print(nums)
        #         return i
        #     elif nums[i] > target:
        #         nums.insert(i, target)
        #         print(nums)
        #         return i
        #     elif nums[len(nums)-1]<target:
        #         nums.append(target)
        #         print(nums)
        #         return nums.index(target)

‘’’
def search_insert(list, target):
left = 0
right = len(list)-1
flag = 0
if (target > list[right]):
return right+1
while(left < right):
i = int(left + (right - left)/2)
if (target > list[i]):
left = i + 1
else:
right = i
flag = left
return flag

‘’’

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