279. Perfect Squares

• Total Accepted: 48616
• Total Submissions: 142008
• Difficulty: Medium

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.

穷举会： Time Limit Exceeded

加入几个分支判断，除去大部分没用分支

public class Solution {
    public int numSquares(int n) {
        if(n == 0){
            return 0;
        }
        int count = 0;
        while (count*count <= n){
            if(count*count == n){
                return 1;
            }
            count++;

        }

        int nums [] = new int[count];
        DFS(n,nums,count-1,0);
        return result;
    }

    int result = Integer.MAX_VALUE;
    private void DFS(int n,int[] nums,int selectNum,int preCount){  // count:平方数最大是几，n：sum target,nums 1——count
        if(n == 0){
            int newResult = getResult(nums);
            if(result > newResult){  //time limit error 优化：加入preCount,统计到目前为止已用的的数量
                result = newResult;
            }
            return;
        }
        if(preCount >= result){
            return;
        }
        if(selectNum >= 1){
            int maxLoop = n / selectNum /selectNum;
            if(preCount+maxLoop > result) return;  //time limit error 优化：加入preCount,统计到目前为止已用的的数量
            for(int i = maxLoop;i >= 0;i--){
                nums[selectNum] = i;
                DFS(n-i*selectNum*selectNum,nums,selectNum-1,preCount+i);
            }
        }
    }

    private int getResult(int[]nums){
        int result = 0;
        for(int i = 1;i < nums.length;i++){
//            System.out.print(nums[i] + " ");
            result+=nums[i];
        }
//        System.out.println("result:"+result);
        return result;
    }
}