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minimumabsolutedifference.go
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minimumabsolutedifference.go
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package algorithm
/*
原题:https://leetcode-cn.com/problems/minimum-absolute-difference/
1200. 最小绝对差
给你个整数数组 arr,其中每个元素都 不相同。
请你找到所有具有最小绝对差的元素对,并且按升序的顺序返回。
示例 1:
输入:arr = [4,2,1,3]
输出:[[1,2],[2,3],[3,4]]
示例 2:
输入:arr = [1,3,6,10,15]
输出:[[1,3]]
示例 3:
输入:arr = [3,8,-10,23,19,-4,-14,27]
输出:[[-14,-10],[19,23],[23,27]]
提示:
2 <= arr.length <= 10^5
-10^6 <= arr[i] <= 10^6
分析:
使用二分法计算,先使用快速排序将数组排序,然后用二分法计算出最短距离,最后遍历数组,找到所有距离为最小值的数值对
*/
func MinimumAbsDifference(arr []int) [][]int {
res := make([][]int, 0)
length := len(arr)
if length < 2 {
return res
}
//排序
quickSort(arr, 0, length-1)
//先计算出最小值
minV := computeDistance(arr, 0, length-1)
//遍历数组,判断距离为minV的数对
for i := 1; i < length; i++ {
if arr[i]-arr[i-1] == minV {
res = append(res, []int{arr[i-1], arr[i]})
}
}
return res
}
//迭代计算距离
func computeDistance(arr []int, startIndex int, endIndex int) int {
maxV := 10000000
if startIndex >= endIndex {
return maxV
}
dv := []int{maxV, maxV, maxV, maxV}
middleIndex := (startIndex + endIndex) / 2
dv[0] = computeDistance(arr, startIndex, middleIndex-1)
dv[1] = computeDistance(arr, middleIndex+1, endIndex)
if middleIndex-1 >= startIndex && arr[middleIndex]-arr[middleIndex-1] < dv[2] {
dv[2] = arr[middleIndex] - arr[middleIndex-1]
}
if middleIndex+1 <= endIndex && arr[middleIndex+1]-arr[middleIndex] < dv[3] {
dv[3] = arr[middleIndex+1] - arr[middleIndex]
}
d := maxV
for _, v := range dv {
if d > v {
d = v
}
}
return d
}