题解 | #丑数#

class Solution {
public:
    int GetUglyNumber_Solution(int index) {
        // 动态规划
        vector<int> dp(index + 1);
        dp[1] = 1;
        int p2 = 1;
        int p3 = 1;
        int p5 = 1;
        for(int i = 2; i <= index; i++) {
            int num2 = dp[p2] * 2, num3 = dp[p3] * 3, num5 = dp[p5] * 5;
            int minN = min(num2, num3);
            minN = min(minN, num5);
            dp[i] = minN;
            // 分别判断当前的最小值与哪一个数相等，排除重复情况
            if(minN == num2) {
                p2++;
            }
            if(minN == num3) {
                p3++;
            }
            if(minN == num5){
                p5++;
            }
        }
        return dp[index];
    }
};