题目链接:点击打开链接
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];
}
};