2021-02-25: Given a positive array arr, please divide all the numbers in arr into two sets. If the length of arr is even, the two sets contain the same number of numbers; if the length of arr is odd, the number of the two sets contains only one difference. Please try to make the cumulative sum of the two sets as close as possible, and return the cumulative sum of the smaller set in the closest case.
Fu Ge's answer 2020-02-25:
Natural wisdom is enough .
1. Recursion. There is code.
2. Dynamic planning. dp is a three-dimensional array. There is code.
The code is written in golang, the code is as follows:
package main
import "fmt"
const INT_MAX = int(^uint(0) >> 1)
const INT_MIN = ^INT_MAX
func main() {
arr := []int{1, 2, 3, 4, 100}
ret := right1(arr)
fmt.Println("1.递归：", ret)
ret = right2(arr)
fmt.Println("2.动态规划：", ret)
}
func right1(arr []int) int {
if len(arr) < 2 {
return 0
}
sum := 0
for _, v := range arr {
sum += v
}
if len(arr)&1 == 0 {
return process1(arr, 0, len(arr)/2, sum/2)
} else {
ret1 := process1(arr, 0, len(arr)/2, sum/2)
ret2 := process1(arr, 0, len(arr)/2+1, sum/2)
if ret1 < ret2 {
return ret2
} else {
return ret1
}
}
}
func process1(arr []int, i int, picks int, rest int) int {
if i == len(arr) {
if picks == 0 {
return 0
} else {
return -1
}
} else {
p1 := process1(arr, i+1, picks, rest)
p2 := -1
next := -1
if arr[i] <= rest {
next = process1(arr, i+1, picks-1, rest-arr[i])
}
if next != -1 {
p2 = arr[i] + next
}
return GetMax(p1, p2)
}
}
func right2(arr []int) int {
if len(arr) < 2 {
return 0
}
sum := 0
for _, v := range arr {
sum += v
}
sum >>= 1
N := len(arr)
M := (len(arr) + 1) >> 1
dp := make([][][]int, N)
for i := 0; i < N; i++ {
dp[i] = make([][]int, M+1)
for j := 0; j < M+1; j++ {
dp[i][j] = make([]int, sum+1)
for k := 0; k < sum+1; k++ {
dp[i][j][k] = INT_MIN
}
}
}
for i := 0; i < N; i++ {
for k := 0; k <= sum; k++ {
dp[i][0][k] = 0
}
}
for k := 0; k <= sum; k++ {
if arr[0] <= k {
dp[0][1][k] = arr[0]
} else {
dp[0][1][k] = INT_MIN
}
}
for i := 1; i < N; i++ {
for j := 1; j <= GetMin(i+1, M); j++ {
for k := 0; k <= sum; k++ {
dp[i][j][k] = dp[i-1][j][k]
if k-arr[i] >= 0 {
dp[i][j][k] = GetMax(dp[i][j][k], dp[i-1][j-1][k-arr[i]]+arr[i])
}
}
}
}
return GetMax(dp[N-1][M][sum], dp[N-1][N-M][sum])
}
func GetMin(a int, b int) int {
if a < b {
return a
} else {
return b
}
}
func GetMax(a int, b int) int {
if a > b {
return a
} else {
return b
}
}The execution results are as follows: