1818. 绝对差值和

题目链接	1818. 绝对差值和
思路	排序+二分
题解链接	运用「二分」找最佳替换方案
关键点	转换为查找最小值delta：对nums1进行排序后，从中二分查找nums2[i]的最接近值（考虑到绝对值，需要检查left & right两个位置）
时间复杂度	\(O(n\log n)\)
空间复杂度	\(O(n)\)

代码实现：

class Solution:
    def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
        total = 0
        sorted_ = sorted(nums1)
        n = len(nums1)
        delta = inf
        for i in range(n):
            diff = abs(nums1[i] - nums2[i])
            total += diff

            left, right = -1, n
            while left + 1 < right:
                mid = (left+right) // 2
                if sorted_[mid] < nums2[i]:
                    left = mid
                else:
                    right = mid

            if left >= 0:
                delta = min(delta, abs(sorted_[left] - nums2[i]) - diff)
            if right < n:
                delta = min(delta, abs(sorted_[right] - nums2[i]) - diff)
        return (total + delta) % (10 ** 9 + 7)

Python-标准库

class Solution:
    def minAbsoluteSumDiff(self, nums1: List[int], nums2: List[int]) -> int:
        total = 0
        sorted_ = sorted(nums1)
        n = len(nums1)
        delta = inf
        for i in range(n):
            diff = abs(nums1[i] - nums2[i])
            total += diff

            right = bisect_left(sorted_, nums2[i])
            left = right-1
            if left >= 0:
                delta = min(delta, abs(sorted_[left] - nums2[i]) - diff)
            if right < n:
                delta = min(delta, abs(sorted_[right] - nums2[i]) - diff)
        return (total + delta) % (10 ** 9 + 7)