摘要:动态规划法建立空数组从到每个数包含最少平方数情况,先所有值为将到范围内所有平方数的值赋两次循环更新,当它本身为平方数时,简化动态规划法四平方和定理法
Problem
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
ExampleGiven n = 12, return 3 because 12 = 4 + 4 + 4
Given n = 13, return 2 because 13 = 4 + 9
这道题在OJ有很多解法,公式法,递归法,动规法,其中公式法时间复杂度最优(four square theorem)。
不过我觉得考点还是在动规吧,也更好理解。
1. 动态规划法
public class Solution { public int numSquares(int n) { //建立空数组dp:从0到n每个数包含最少平方数情况,先fill所有值为Integer.MAX_VALUE; int[] dp = new int[n+1]; Arrays.fill(dp, Integer.MAX_VALUE); //将0到n范围内所有平方数的dp值赋1; for (int i = 0; i*i <= n; i++) { dp[i*i] = 1; } //两次循环更新dp[i+j*j],当它本身为平方数时,dp[i+j*j] < dp[i]+1 for (int i = 0; i <= n; i++) { for (int j = 0; i+j*j <= n; j++) { dp[i+j*j] = Math.min(dp[i]+1, dp[i+j*j]); } } return dp[n]; } }
2. 简化动态规划法
public class Solution { public int numSquares(int n) { int[] dp = new int[n+1]; for (int i = 0; i <= n; i++) { dp[i] = i; for (int j = 0; j*j <= i; j++) { dp[i] = Math.min(dp[i], dp[i-j*j]+1); } } return dp[n]; } }
3. 四平方和定理法
public class Solution { public int numSquares (int n) { int m = n; while (m % 4 == 0) m = m >> 2; if (m % 8 == 7) return 4; int sqrtOfn = (int) Math.sqrt(n); if (sqrtOfn * sqrtOfn == n) //Is it a Perfect square? return 1; else { for (int i = 1; i <= sqrtOfn; ++i){ int remainder = n - i*i; int sqrtOfNum = (int) Math.sqrt(remainder); if (sqrtOfNum * sqrtOfNum == remainder) return 2; } } return 3; } }
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