Perfect Squares:
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:
#include <iostream>
#include <vector>
#include <queue>
#include <stdexcept>
using namespace std;
class Solution {
public:
int numSquares(int n) {
assert( n > 0 );
queue< pair< int, int > > q; // 第一个int表示:具体是哪一个数字。第二个int表示:建立的图中经历了几段路径走到了这个数字
q.push( make_pair( n, 0 ));
vector<bool> visited( n+1, false); // n+1个结点是否被访问过,初始化都为false
visited[n] = true;
while ( !q.empty() ){
int num = q.front().first;
int step = q.front().second;
q.pop();
// if ( num == 0 )
// return step;
for( int i = 1; ; i++){
int a = num - i * i;
if ( a < 0)
break;
if ( a == 0)
return step + 1;
if( ! visited[a] ) {
q.push(make_pair(a, step + 1));
visited[a] = true;
}
}
}
throw invalid_argument("No Solution.");
}
};
总结: 此题可以转化为图中广度优先搜索求最短路径,具体建模思想见图:
举个例子看一下:
这个建模思想要学会!