和为s的连续正数序列

题目描述

小明很喜欢数学,有一天他在做数学作业时,要求计算出9~16的和,他马上就写出了正确答案是100。但是他并不满足于此,他在想究竟有多少种连续的正数序列的和为100(至少包括两个数)。没多久,他就得到另一组连续正数和为100的序列:18,19,20,21,22。现在把问题交给你,你能不能也很快的找出所有和为S的连续正数序列? Good Luck!

输出描述:

输出所有和为S的连续正数序列。序列内按照从小至大的顺序，序列间按照开始数字从小到大的顺序

思路一

import java.util.ArrayList;

public class Continuous {
    public static void main(String[] args){

        Continuous continuous = new Continuous();
        ArrayList<ArrayList<Integer>> arrayLists = continuous.FindContinuousSequence(9);
        for(ArrayList<Integer> sub : arrayLists){
            for(int i = 0; i < sub.size(); i++){
                System.out.print(sub.get(i));

            }
            System.out.println(sub.toArray());
        }
    }

    public ArrayList<ArrayList<Integer> > FindContinuousSequence(int sum) {
        ArrayList<ArrayList<Integer> > result = new ArrayList<>();
        int low = 1,high = 2;
        while(high > low){
            //由于是连续的，差为1的一个序列，那么求和公式是(a0+an)*n/2
            int cur = (high + low) * (high - low + 1) / 2;
            if(cur == sum){
                ArrayList<Integer> list = new ArrayList<>();
                for(int i=low;i<=high;i++){
                    list.add(i);
                }
                result.add(list);
                high++;
            }else if(cur < sum){
                high++;
            }else{
                low++;
            }
        }
        return result;
    }
}

思路二

import org.omg.CORBA.PUBLIC_MEMBER;

import javax.swing.text.StyleContext;
import java.util.ArrayList;

public class COtinu {
    public static void main(String[] args){
        COtinu cOtinu = new COtinu();
        ArrayList<ArrayList<Integer>> arrayLists = cOtinu.FindContinuousSequence(9);
        for(ArrayList<Integer> sub : arrayLists){
            for(int i = 0; i < sub.size(); i++){
                System.out.print(sub.get(i));

            }
            System.out.println(sub.toArray());
        }
    }


    public ArrayList<Integer> addToSubList(int low,int high){
        ArrayList<Integer> subList = new ArrayList<>();
        for(;low <= high;low++){
            subList.add(low);
        }
        return subList;
    }

    public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum) {
        ArrayList<ArrayList<Integer>> list = new ArrayList<>();


        if(sum == 1){
            ArrayList<Integer> subList = new ArrayList<>();
            subList.add(1);
            list.add(subList);
            return list;
        }
        if(sum == 2) return  null;

        int low = 1;
        int high = 2;
        int middle = (sum - 1) / 2;
        int curSum = low + high;

        while (low < high){
            if(curSum == sum){
                list.add(addToSubList(low, high));
            }
            while (curSum > sum && low < middle){
                curSum -= low;
                low++;
                if(curSum == sum){
                    list.add(addToSubList(low, high));
                }
            }
            high++;
            curSum += high;
        }

        return list;

    }
}

该种思路的时间复杂度过大

思路三

public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum) {
        ArrayList<ArrayList<Integer>> ans = new ArrayList<>();
        for (int n = (int) Math.sqrt(2 * sum); n >= 2; n--) {
            //n分别为奇数或偶数的情况
            //n为奇数时需要满足(n & 1) == 1 && sum % n == 0
            //n为偶数时需要满足(sum % n) * 2 == n)
            if ((n & 1) == 1 && sum % n == 0 || (sum % n) * 2 == n) {
                ArrayList<Integer> list = new ArrayList<>();
                for (int j = 0, k = (sum / n) - (n - 1) / 2; j < n; j++, k++) {
                    list.add(k);
                }
                ans.add(list);
            }
        }
        return ans;
    }