JS带库存的抽奖算法问题

在仅有总数和剩余库存的情况下,如何动态的调整抽奖概率呢,比如一等奖如果剩余库存为0了,之后就不会再有一等奖出现了

[
    {
        name: '一等奖',
        total: 1,
        remain: 1
    },
    {
        name: '二等奖',
        total: 2,
        remain: 2
    },
    {
        name: '三等奖',
        total: 10,
        remain: 10
    },
    {
        name: '四等奖',
        total: 20,
        remain: 20
    },
]
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1 个回答

大概思路应该是这样子,不知道是不是你想要的

let lotteryAwardList=[
    {
        name: '一等奖',
        total: 1,
        remain: 1
    },
    {
        name: '二等奖',
        total: 2,
        remain: 2
    },
    {
        name: '三等奖',
        total: 10,
        remain: 10
    },
    {
        name: '四等奖',
        total: 20,
        remain: 20
    },
]
function lottery(){
    let chooseAwardName=''
    if(lotteryAwardList.length>0){
        let total=0,arr=[],copylotteryAwardList=[];
        lotteryAwardList.map((award,index)=>{
            total+=award.remain;
            if(award.remain>0){
                copylotteryAwardList.push(award)
                 for(var i=0;i<award.remain;i++){
                    arr.push(index)
                }
            }
           
        })
        let randomNum = parseInt( Math.random() * arr.length );
        let chooseAwardIndex=arr[randomNum]
        chooseAwardName=copylotteryAwardList[chooseAwardIndex].name
        lotteryAwardList[chooseAwardIndex].remain=lotteryAwardList[chooseAwardIndex].remain-1;
        if(copylotteryAwardList[chooseAwardIndex].remain===0){
            lotteryAwardList.splice(chooseAwardIndex,1)
        }
    }else{
        chooseAwardName='无'
    }
    
    return chooseAwardName
}
lottery()
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