Combination Sum II 无序数组中找组合(每个元素只能用一次)使得和为target@LeetCode@LeetCode
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和上一题思想一样,区别在于每个元素只能用一次,所以start的位置也要改变。
另外一个问题是如何处理返回大集合中有重复子集合的问题。最容易想到的就是用HashSet来过滤一遍,参考了http://blog.csdn.net/u011095253/article/details/9158423
的大作后发现有一种更绿色的方法,就是多用一个while循环跳过相同的元素!这个技巧在很多情况下都很好用的!
package Level4;import java.util.ArrayList;import java.util.Arrays;/** * Combination Sum II * * Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.Each number in C may only be used once in the combination.Note:All numbers (including target) will be positive integers.Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).The solution set must not contain duplicate combinations.For example, given candidate set 10,1,2,7,6,1,5 and target 8, A solution set is: [1, 7] [1, 2, 5] [2, 6] [1, 1, 6] * */public class S40 {public static void main(String[] args) {int[] num = {10,1,2,7,6,1,5};int target = 8;System.out.println(combinationSum2(num, target));}public static ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) { ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>(); ArrayList<Integer> al = new ArrayList<Integer>(); Arrays.sort(num); dfs(num, target, 0, ret, al); return ret; }public static void dfs(int[] num, int remain, int start, ArrayList<ArrayList<Integer>> ret, ArrayList<Integer> al){if(remain == 0){ret.add(new ArrayList<Integer>(al));return;}for(int i=start; i<num.length; i++){int rest = remain - num[i];if(rest < 0){return;}al.add(num[i]);dfs(num, rest, i+1, ret, al);// 因为不能重复使用元素,所以每次用完后就跳过al.remove(al.size()-1);// 跳过相同的元素while(i<num.length-1 && num[i]==num[i+1]){i++;}}}}
注意:
while(i+1<num.length && num[i+1]==num[i]){
i++;
}
这里i只能和i+1比较,因为i是从0开始的,所以如果与i-1比较则会因为越界而无法继续比较。
public class Solution { public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) { ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>(); ArrayList<Integer> al = new ArrayList<Integer>(); Arrays.sort(num); dfs(num, target, 0, al, ret); return ret; } public void dfs(int[] num, int target, int pos, ArrayList<Integer> al, ArrayList<ArrayList<Integer>> ret){ if(target < 0){ return; } if(target == 0){ ret.add(new ArrayList<Integer>(al)); return; } for(int i=pos; i<num.length; i++){ al.add(num[i]); dfs(num, target-num[i], i+1, al, ret); al.remove(al.size()-1); while(i+1<num.length && num[i+1]==num[i]){ i++; } } }}
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