求数组中和为给定值的所有子序列

package com.zxt.algorithm;import java.util.ArrayList;import java.util.Arrays;/** * 求数组中和为给定值的所有子序列 *  * 如：数组[1,2,3,4,5,6]，sum=7时，满足条件的子数组有[1,2,4],[3,4],[2,5],[1,6]; *  * @author zxt * */public class FixedSumSequence {public static void main(String[] args) {int[] arr = { 3, 2, 5, 5, 5, 6, 1, 4, 4 };// 数组升序排序Arrays.sort(arr);ArrayList<ArrayList<Integer>> lists = new ArrayList<ArrayList<Integer>>();boolean[] visited = new boolean[arr.length];// 求和为sum的所有组合数backTrack(arr, visited, 0, 0, 7, lists);System.out.println(lists);System.out.println(lists.size());}// 求一个数组中和为定值的所有子序列public static void backTrack(int[] input, boolean[] visited, int n, int sum, int key,ArrayList<ArrayList<Integer>> lists) {// 如果条件为：if (sum % key == 0 && sum != 0) 则是求所有和为定值整数倍的序列if (sum == key) {ArrayList<Integer> list = new ArrayList<Integer>();for (int j = 0; j <= n; j++) {if (visited[j]) {list.add(input[j]);}}lists.add(list);return;}for (int i = n; i < input.length; i++) {if (!visited[i]) {sum += input[i];visited[i] = true;if (i != input.length - 1) {backTrack(input, visited, i + 1, sum, key, lists);} else {backTrack(input, visited, i, sum, key, lists);}visited[i] = false;sum -= input[i];// 重复的元素只统计一次，否则会出现重复（而次操作的前提是，序列已经排序。所以参数input数组是排序好的）while (i < input.length - 1 && input[i] == input[i + 1]) {i++;}}}}}