题解 | #左旋转字符串#

// 求第i个丑数，就是找前i-2个丑数乘以2，3，5里大于第i-1个丑数的最小值
function GetUglyNumber_Solution(index) {
    if (index <= 0) {
        return 0;
    }
    let arr = [1];
    let i2=0,i3=0,i5=0  
    let cur = 0;
    while (arr.length < index) {
        arr.push(Math.min(arr[i2] * 2, arr[i3] * 3, arr[i5] * 5));//尾部添加丑数
        const current = arr[arr.length - 1];
        while (arr[i2] * 2 <= current) {
            i2++; // 更新i2索引（丑数的性质）
        }
        while (arr[i3] * 3 <= current) {
            i3++;// 更新i3索引
        }
        while (arr[i5] * 5 <= current) {
            i5++;// 更新i5索引
        }
    }
    return arr[index - 1];
}
module.exports = {
    GetUglyNumber_Solution : GetUglyNumber_Solution
};