279. Perfect Squares （DFS, DP）

https://leetcode.com/problems/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.

分析：

分解成完全平方数，可以把分解的过程看成寻找路径。两个数值之间相差的是完全平方数就可以连通。

第一种方法是使用BFS， 这种思路要注意的是前面到达的肯定路径更短，所以设置一个访问标记，前面访问过就加到队列并且设置标记为访问过。

public int numSquares(int n) {
        List<Integer> squares = generateSquares(n);
        boolean[] flag = new boolean[n+1];
        flag[n] = true;
        Queue<Integer> queue = new LinkedList<Integer>();
        queue.add(n);
        int num = 0;
        while(!queue.isEmpty()){
            int size = queue.size();
            num++;
            while(size-->0){
                int cur = queue.poll();
                for(int s : squares){
                    int rest = cur - s;
                    if(rest<0)
                        break;
                    if(rest==0)
                        return num;
                    if(flag[rest])
                        continue;
                    flag[rest] = true;//BFS先到达的肯定路径更短，标记过就不用考虑
                    queue.add(rest);                        
                }
            }
        }
        return num;
    }
    
    private List<Integer> generateSquares(int n){
        List<Integer> res = new ArrayList<Integer>();
        int square = 1, diff=3;
        while(square<=n){
            res.add(square);
            square += diff;
            diff += 2;
        }
        return res;
    }

还有一种方法是DP。 

dp[0] = 0 
dp[1] = dp[0]+1 = 1
dp[2] = dp[1]+1 = 2
dp[3] = dp[2]+1 = 3
dp[4] = Min{ dp[4-1*1]+1, dp[4-2*2]+1 } 
      = Min{ dp[3]+1, dp[0]+1 } 
      = 1				
dp[5] = Min{ dp[5-1*1]+1, dp[5-2*2]+1 } 
      = Min{ dp[4]+1, dp[1]+1 } 
      = 2
						.
						.
						.
dp[13] = Min{ dp[13-1*1]+1, dp[13-2*2]+1, dp[13-3*3]+1 } 
       = Min{ dp[12]+1, dp[9]+1, dp[4]+1 } 
       = 2
						.
						.
						.
dp[n] = Min{ dp[n - i*i] + 1 },  n - i*i >=0 && i >= 1

public int numSquares(int n) {
        int[] dp = new int[n+1];
        Arrays.fill(dp, Integer.MAX_VALUE);
        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];
    }