Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.
For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
动态规划。
class Solution {
public int numSquares(int n) {
int[] dp = new int[n+1];
dp[0]=0;
for(int i=1;i<=n;i++){
int min = Integer.MAX_VALUE;
int j = 1;
while(i-j*j>=0){
min = Math.min(min,dp[i-j*j]+1);
j++;
}
dp[i] = min;
}
return dp[n];
}
}
public int numSquares(int n) {
int[] dp = new int[n+1];
dp[0]=0;
for(int i=1;i<=n;i++){
int min = Integer.MAX_VALUE;
int j = 1;
while(i-j*j>=0){
min = Math.min(min,dp[i-j*j]+1);
j++;
}
dp[i] = min;
}
return dp[n];
}
}
本文介绍了一种使用动态规划解决的问题:给定一个正整数n,找到最少数量的完全平方数(如1, 4, 9, 16...),这些数相加等于n。例如,当n=12时,返回3(因为12=4+4+4);当n=13时,返回2(因为13=4+9)。通过动态规划的方法,我们能够有效地找到这个问题的解决方案。
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