算法积累001
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