需求
公司有一个比较坑爹的报销方案，需要根据一堆碎的发票中，凑出一个目标金额，要求误差在1块钱以内
缺点：每次人肉去对比吧，浪费大量的时间
解决
下面贴出golang实现的方案
package main
import (
"fmt"
"strconv"
"time"
)
type InvoiceCounter struct {
maxValue int //期望值（单元为分）
items []int //发票金额（单元为分）
overflow int //允许的误差值（单元为分）
}
//items:所有发票 maxValue:目标金额 overflow:允许误差金额
func NewInvoiceCounter(items []float64, maxValue float64, overflow float64) *InvoiceCounter {
obj := &InvoiceCounter{}
obj.maxValue = obj.dollarToCent(maxValue)
obj.overflow = obj.dollarToCent(overflow)
centItems := make([]int, len(items))
for i, v := range items {
centItems[i] = obj.dollarToCent(v)
}
obj.items = centItems
return obj
}
//元转分
func (this *InvoiceCounter) dollarToCent(value float64) int {
value, _ = strconv.ParseFloat(fmt.Sprintf("%.2f", value), 64)
return int(value * 100)
}
//分转元
func (this *InvoiceCounter) centToDollar(v int) float64 {
value := float64(v)
value, _ = strconv.ParseFloat(fmt.Sprintf("%.2f", value/100), 64)
return value
}
//执行计算，返回所有方案
func (this *InvoiceCounter) Run() [][]float64 {
items := this.items
n := len(this.items)
max := this.maxValue + this.overflow
states := this.createStates(len(this.items), max+1)
states[0][0] = true
if items[0] <= max {
states[0][items[0]] = true
}
for i := 1; i < n; i++ {
//不选
for j := 0; j <= max; j++ {
if states[i-1][j] {
states[i][j] = states[i-1][j]
}
}
//选中
for j := 0; j <= max-items[i]; j++ {
if states[i-1][j] {
states[i][j+items[i]] = true
}
}
}
//获取最终所有满足的方案
res := make([][]float64, 0)
for j := this.maxValue; j <= max; j++ {
for i := 0; i < n; i++ {
if states[i][j] {
//判断必须最后一个选中才算，要不区间有重合 比如前5个元素已经满足目标金额了，state[5][w]=true，然后state[6][w]也是true，存在重复的方案
if j-items[i] >= 0 && states[i-1][j-items[i]] == true {
res = append(res, this.getSelected(states, items, i, j))
}
}
}
}
return res
}
//获取所有选中的元素（倒推）
func (this *InvoiceCounter) getSelected(states [][]bool, items []int, n, max int) []float64 {
var selected = make([]int, 0)
for i := n; i >= 1; i-- {
//元素被选中
if max-items[i] >= 0 && states[i-1][max-items[i]] == true {
selected = append([]int{items[i]}, selected...)
max = max - items[i]
} else {
//没选，max重量不变，直接进入下一次
}
}
if max != 0 {
selected = append([]int{items[0]}, selected...)
}
dollarItems := make([]float64, len(selected))
for i, v := range selected {
dollarItems[i] = this.centToDollar(v)
}
return dollarItems
}
//初始化所有状态
func (this *InvoiceCounter) createStates(n, max int) [][]bool {
states := make([][]bool, n)
for i, _ := range states {
states[i] = make([]bool, max)
}
return states
}使用示例
func main() {
//所有发票金额
items := []float64{100, 101, 103, 105, 106, 132, 129, 292, 182, 188, 224.3, 40.5, 35.9, 32.5, 39, 12, 17.5, 28, 35, 34, 26.32, 28, 35, 39, 25, 1, 24, 35, 45, 47, 32.11, 45, 32, 38.88, 44, 36.5, 35.8, 45, 26.5, 33, 25, 364, 27.3, 39.2, 180, 279, 282, 281, 285, 275, 277, 278, 200, 201, 1959.12, 929.53, 1037.03, 1033.9}
//目标金额5000钱，允许差额1块钱
obj := NewInvoiceCounter(items, 5000, 1)
startTime := time.Now()
//执行计算，返回所有结果方案
res := obj.Run()
//打印所有方案
for _, v := range res {
fmt.Println(v)
}
fmt.Printf("total:%d used time：%s\n", len(res), time.Now().Sub(startTime))
}
运行结果
[100 101 103 105 106 132 129 292 182 188 224.3 40.5 12 17.5 35 34 26.32 28 35 39 25 1 24 35 45 47 45 32 38.88 44 36.5 45 26.5 33 25 364 27.3 39.2 180 279 282 281 285 275 277 278]
[100 101 103 105 132 129 292 182 188 35.9 39 12 17.5 28 35 34 26.32 28 35 39 25 1 24 35 45 47 32.11 45 32 38.88 44 36.5 35.79 45 26.5 33 25 364 27.3 39.2 180 279 282 281 285 275 277 278 200]
...
[35.9 25 24 38.88 36.5 35.79 45 26.5 33 25 27.3 39.2 180 279 282 281 285 275 277 278 200 201 1037.03 1033.9]
total:577 used time：97.048224ms耗时97毫秒，共计算出577种方案，就是这么爽！