LeetCode（算法）- 264. 丑数 II

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题目大意：略

解题思路：略

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AC 代码

• Java

// 枚举暴力（超时）
class Solution {
    public int nthUglyNumber(int n) {
        int num = 1, res = -1;
        for (int i = 0; i < n;) {
            if (isUglyNum(num)) {
                res = num;
                i++;
            }
            num++;
        }
        return res;
    }
 
    private boolean isUglyNum(int num) {
        while (num % 2 == 0) {
            num /= 2;
        }
 
        while (num % 3 == 0) {
            num /= 3;
        }
 
        while (num % 5 == 0) {
            num /= 5;
        }
 
        return num == 1;
    }
}
 
// AC
class Solution {
    public int nthUglyNumber(int n) {
        int a = 0, b = 0, c = 0;
        int[] dp = new int[n];
        dp[0] = 1;
        for(int i = 1; i < n; i++) {
            int n2 = dp[a] * 2, n3 = dp[b] * 3, n5 = dp[c] * 5;
            dp[i] = Math.min(Math.min(n2, n3), n5);
            if(dp[i] == n2) a++;
            if(dp[i] == n3) b++;
            if(dp[i] == n5) c++;
        }
        return dp[n - 1];
    }
}

• C++

class Solution {
public:
    int nthUglyNumber(int n) {
        int a = 0, b = 0, c = 0;
        int dp[n];
        dp[0] = 1;
        for(int i = 1; i < n; i++) {
            int n2 = dp[a] * 2, n3 = dp[b] * 3, n5 = dp[c] * 5;
            dp[i] = min(min(n2, n3), n5);
            if(dp[i] == n2) a++;
            if(dp[i] == n3) b++;
            if(dp[i] == n5) c++;
        }
        return dp[n - 1];
    }
};