📅  最后修改于: 2023-12-03 15:22:04.383000             🧑  作者: Mango
在软件开发的过程中,经常会有需要将两个数组的元素相同的情况。但是,在实际开发中,两个数组的元素往往不完全相同,需要通过增加或减少元素来使得两个数组的元素相等。如何实现以最小增量或减量使两个数组的元素相同呢?下面将介绍几种实现方法。
代码实现:
func minIncrementForEqualArrays(arr1 []int, arr2 []int) int {
sort.Ints(arr1)
sort.Ints(arr2)
i, j, count := 0, 0, 0
for i < len(arr1) && j < len(arr2) {
if arr1[i] == arr2[j] {
i++
j++
} else if arr1[i] < arr2[j] {
arr1[i] = arr1[i] + 1
count++
i++
} else {
arr2[j] = arr2[j] + 1
count++
j++
}
}
for i < len(arr1) {
arr1[i] = arr1[i] + 1
count++
i++
}
for j < len(arr2) {
arr2[j] = arr2[j] + 1
count++
j++
}
return count
}
代码实现:
func minIncrementForEqualArrays(arr1 []int, arr2 []int) int {
m1 := make(map[int]int)
m2 := make(map[int]int)
for _, num := range arr1 {
m1[num]++
}
for _, num := range arr2 {
m2[num]++
}
count := 0
for num, c1 := range m1 {
if c2, ok := m2[num]; ok {
delete(m1, num)
delete(m2, num)
count += min(c1, c2)
}
}
sum1, sum2 := 0, 0
for num, c := range m1 {
sum1 += num * c
}
for num, c := range m2 {
sum2 += num * c
}
if sum1 < sum2 {
m1, m2 = m2, m1
}
for num, c := range m1 {
for i := 0; i < c; i++ {
for {
if _, ok := m2[num+1]; !ok {
m2[num+1] = 1
count += num + 1 - num
break
}
num++
m2[num]++
}
}
}
return count
}
代码实现:
type unionFind struct {
parent, size []int
count int
}
func newUnionFind(n int) *unionFind {
parent := make([]int, n)
size := make([]int, n)
for i := range parent {
parent[i] = i
size[i] = 1
}
return &unionFind{parent, size, n}
}
func (uf *unionFind) find(x int) int {
if x != uf.parent[x] {
uf.parent[x] = uf.find(uf.parent[x])
}
return uf.parent[x]
}
func (uf *unionFind) merge(x, y int) {
rootX, rootY := uf.find(x), uf.find(y)
if rootX == rootY {
return
}
if uf.size[rootX] < uf.size[rootY] {
rootX, rootY = rootY, rootX
}
uf.parent[rootY] = rootX
uf.size[rootX] += uf.size[rootY]
uf.count--
}
func minIncrementForEqualArrays(arr1 []int, arr2 []int) int {
sort.Ints(arr1)
sort.Ints(arr2)
uf := newUnionFind(len(arr1) + len(arr2))
for i := range arr1 {
for j := range arr2 {
if arr1[i] == arr2[j] {
uf.merge(i, j+len(arr1))
}
}
}
count := 0
for i := range arr1 {
if uf.find(i) == i {
size := uf.size[i]
count += size - 1
uf.size[i] = 1
uf.parent[i] = uf.find(i + 1)
}
}
for i := range arr2 {
if uf.find(i+len(arr1)) == i+len(arr1) {
size := uf.size[i+len(arr1)]
count += size - 1
uf.size[i+len(arr1)] = 1
uf.parent[i+len(arr1)] = uf.find(i+len(arr1) + 1)
}
}
return count
}
以上是三种实现方法。对于每种方法,都有其适用的场景,需要根据实际情况灵活应用。