迅速找出数组a中是否存在相加后等于整数x的两个元素

public class QuickSearch {
  
    public static int getMiddle(List<Integer> list, int low, int high) {
          Integer temp = list.get(high);   
            while (low < high) { 
                while (low < high && list.get(low) < temp) { 
                    low++; 
                } 
                list.set(high, list.get(low));
                while (low < high && list.get(high) > temp) { 
                    high--; 
                } 
                list.set(low, list.get(high));
                if (list.get(high) == temp) {
                    break;
                }
            } 
            list.set(high, temp);
        return low;                  
    }
     
    //先定位3个点
    public static void quickSearch(List<Integer> list, int low, int high) { 
        if (low < high) { 
            int target = getMiddle(list, low, high); 
            if (target<2) {
                return;
            }
            int middle = getMiddle(list, low, target-1);
            int lwoMiddle = 0;
            if (middle>0) {
                lwoMiddle = getMiddle(list, low, middle-1);
            }
            int highMiddle = target-1;
            if(middle+1 != target){
                highMiddle = getMiddle(list, middle+1, target-1);
            }
            int lowLow = low;
            int lowHigh = target-1;
            int highLow = target+1;
            int highHigh = high;
            numSearch(list,target,lwoMiddle,highMiddle,lowLow,lowHigh,highLow,highHigh);
        } 
    } 
     
    //找出匹配的相加项
    public static void numSearch(List<Integer> list,int target,int lwoMiddle,int highMiddle,int lowLow, int lowHigh,int highLow,int highHigh){
        Integer targetNum = list.get(target);
        Integer lwoMiddleNum = list.get(lwoMiddle);
        Integer highMiddleNum = list.get(highMiddle);
        boolean flag = true;
        while(flag){
            if (lwoMiddleNum + highMiddleNum > targetNum) {
                recursionSearch(list,lwoMiddle+1,lowHigh,highLow,highMiddle-1,target);
                recursionSearch(list,lowLow,lwoMiddle-1,highMiddle+1,highHigh,target);
                recursionSearch(list,lowLow,lwoMiddle-1,highLow,highMiddle-1,target);
            }else if (lwoMiddleNum + highMiddleNum < targetNum) {
                recursionSearch(list,lwoMiddle+1,lowHigh,highMiddle+1,highHigh,target);
            } else{
                System.out.println(lwoMiddleNum+" + "+highMiddleNum+" = "+targetNum);
                flag = false;
            }
        }
    }
     
    public static void recursionSearch(List<Integer> list, int lowLow, int lowHigh,int highLow,int highHigh,int target){
        int lwoMiddle = -1;
        int highMiddle = -1;
        if (lowLow < lowHigh) {
            lwoMiddle = getMiddle(list, lowLow, lowHigh);   
        }
        if (highLow < highHigh) { 
            highMiddle = getMiddle(list, highLow, highHigh);
        }
        if (lwoMiddle == -1 || highMiddle == -1 ) {
            return;
        }
        numSearch(list,target,lwoMiddle,highMiddle, lowLow, lowHigh, highLow, highHigh);
    }
     
     
    public static void quick(List<Integer> list,int search) { 
        list.add(search);
        if (list.size() > 0) {  
            quickSearch(list, 0, list.size() - 1); 
        } 
    } 
    public static void main(String[] args) {
        List<Integer> list = new ArrayList<Integer>();
        list.add(14);
        list.add(3);
        list.add(12);
        list.add(1);
        list.add(9);
        list.add(7);
        list.add(11);
        list.add(10);
        list.add(5);
        list.add(6);
        quick(list, 10);
    }
}

复杂度：N+(N/2)lg(N/2)。