package com.bleum;

/**
 * 
 * @author aaron.hu
 * 一个排好序的数组，找出两数之和为m的所有组合
 */
public class AlgorithmSample001 {
	public  Integer[] arr = new Integer[]{1,2,3,8,10,13,19,22,23,30,32,34,35,36,37,40,42,44,46,47,50,55};
	int count=0, sum = 56;
	//propose1设置一个中间变量，如果有满足条件的i,j，则后面的遍历的范围会大于a[i]并且小于a[j]
	public void launch(){
		count=0;
		int j=arr.length-1;
		int flag = arr.length-1;
		for(int i=0; i<flag; i++){
			for(j=flag; j>i; j--){
				count++;
				if(arr[i]+arr[j]==sum){
					System.out.println("arr[i]="+arr[i]+",arr[j]="+arr[j]);
					flag = --j;
					break;
				}
				
			}
		}
		System.out.println("Count="+count);
	}
	//propose2卡住一个端，左端只有小于M/2，右端大于M/2才会判断  
	public void launch2(){
		count=0;
		for (int i = 0; i < arr.length; i++) {  
            if (arr[i] <= 56 / 2) {  
                    for (int j = i + 1; j < arr.length; j++) {  
                        if (arr[j] >= 56 / 2) {  
                            count++;  
                            if (arr[i] + arr[j] == 56) {  
                                System.out.println(arr[i] + ":" + arr[j]);  
                            }  
                    }  
  
                }  
            }  
        } 
		System.out.println("Count="+count);
	}
	
}

 Result:

arr[i]=1,arr[j]=55

arr[i]=10,arr[j]=46

arr[i]=19,arr[j]=37

arr[i]=22,arr[j]=34

Count=80

1:55

10:46

19:37

22:34

Count=117