LeetCode 658：寻找K个最接近元素中comp变量未按预期更新

问题描述 #

我正在解决LeetCode第658题「寻找K个最接近元素」，题目要求从给定数组arr中返回长度为k的子数组，包含k个与给定值x最接近的元素。我已完成所有边界条件和约束判断，但在比较逻辑部分遇到问题：

我创建了空列表a，当a的长度不等于k时，遍历数组并比较abs(i - x) < abs(comp - x)，comp初始值为arr[0]，预期当比较成立时将comp更新为i，但实际该变量未按预期更新。测试用例为：arr = [1,1,1,10,10,10]，k = 4，x = 9。

重点排查的代码片段 #

a = []
comp = arr[0]
iteration = 0
i_index = 0
while len(a) != k:
    for i in arr:
        comp_1 = abs(i - x)
        comp_2 = abs(comp - x)
        if comp_1 < comp_2:
            comp == i
            print(f"comp: {comp}")
    arr.pop(arr.index(comp))
    a.append(comp)
return a

完整代码 #

def findClosestElements(self, arr: List[int], k: int, x: int) -> List[int]:
    
    # Constraints
    if k < 1 or k > len(arr):
        return "k must be greater than 0 and less than the arr length"
    if len(arr) < 1 or len(arr) > 10**4:
        return "arr length must be greater than 0 and less than 10^4"
    if x > 10**4:
        return "x must be less than 10^4"
    if sorted(arr) != arr:
        return "arr must be sorted"
    for i in arr:
        if i < -10**4:
            return "arr item cannot be less than -10^4"
    
    
    #Variables 1
    begin = arr[:k]
    end = arr[-k:]
    
    
    # Base cases
    if len(arr) == k:
        return arr
    
    if x < arr[0]:
        return begin
    elif x > arr[-1]:
        return end
    
    try:
        x_index = arr.index(x)
        half_k = int(k/2)   
    #if k == x and x_index != None:
     #   return [x]
        # Captures all other lists that begin at arr[0] or end at arr[-1]
        if x_index - half_k < 0:
            return begin
        elif x_index + half_k > len(arr) - 1:
            return end
        # Create list out of interior of arr if necessary
        else:
            return arr[x_index - half_k : x_index + half_k]
   
    # Means x is not in arr
    except ValueError:
        a = []
        comp = arr[0]
        iteration = 0
        i_index = 0
        while len(a) != k:
            for i in arr:
                print(f"{iteration} - {i_index}:")
                print(f"i: {i}")
                print(f"comp_1: {abs(i - x)}")
                print(f"comp_2: {abs(comp - x)}")
                comp_1 = abs(i - x)
                comp_2 = abs(comp - x)
                if comp_1 < comp_2:
                    comp == i
                    print(f"comp: {comp}")
                i_index += 1
                print("\n")
            iteration += 1
            arr.pop(arr.index(comp))
            a.append(comp)
        return a

内容的提问来源于stack exchange，提问作者Stoutish_goat