Discription
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.
Solution
int numSquares(int n) { //BFS
int sn = sqrt(n);
if (sn*sn == n)
return 1;
vector<int> square, state(n + 1, 0);
queue<int> que;
for (int i = 1; i <= sn; i++){
square.push_back(i*i);
que.push(i*i);
state[i*i] = 1;
}
while (!que.empty()){
int x = que.front();
que.pop();
for (int i : square){
if (x + i <= n && state[x + i] == 0){
state[x + i] = state[x] + 1;
que.push(x + i);
}
}
}
return state[n];
}
int numSquares(int n) { //DP
vector<int> dp(n+1,0);
for (int i = 1; i <= n; i++){
int square = INT_MAX;
for (int j = 1; j*j <= i; j++)
square = min(square, dp[i - j*j] + 1);
dp[i] = square;
}
return dp[n];
}GitHub-Leetcode:
https://github.com/wenwu313/LeetCode
