leetcode264 Ugly Number II
2017-08-28 23:52
405 查看
思路:
dp。
实现:
dp。
实现:
1 class Solution 2 { 3 public: 4 int nthUglyNumber(int n) 5 { 6 vector<int> dp(n); 7 dp[0] = 1; 8 int p2 = 0, p3 = 0, p5 = 0; 9 for (int i = 1; i < n; i++) 10 { 11 dp[i] = min(dp[p2] * 2, min(dp[p3] * 3, dp[p5] * 5)); 12 if (dp[i] == dp[p2] * 2) p2++; 13 if (dp[i] == dp[p3] * 3) p3++; 14 if (dp[i] == dp[p5] * 5) p5++; 15 } 16 return dp[n - 1]; 17 } 18 };
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