title
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, …) which sum to n.
Example 1:
Input: n = 12
Output: 3
Explanation: 12 = 4 + 4 + 4.
Example 2:
Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.
solution
这是bfs回答，还可以dp的
class Solution {
public:
int numSquares(int n) {
queue<int> q;
q.push(n);
int level = 0;
while (!q.empty()) {
int size = q.size();
level++;
while (size--) {
int p = q.front();
q.pop();
for (int i = 1; i <= sqrt(p); i++) {
if (p == i * i) {
return level;
}
else if (p > i*i) {
q.push(p - i * i);
}
}
}
}
return n;
}
};
DP
public int numSquares(int n) {
List<Integer> squareList = generateSquareList(n);
int[] dp = new int[n + 1];
for (int i = 1; i <= n; i++) {
int min = Integer.MAX_VALUE;
for (int square : squareList) {
if (square > i) {
break;
}
min = Math.min(min, dp[i - square] + 1);
}
dp[i] = min;
}
return dp[n];
}
private List<Integer> generateSquareList(int n) {
List<Integer> squareList = new ArrayList<>();
int diff = 3;
int square = 1;
while (square <= n) {
squareList.add(square);
square += diff;
diff += 2;
}
return squareList;
}