Posts Tagged 【dfs】 Combinations

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Combinations

Total Accepted: 38686 Total Submissions: 126278My Submissions

Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

For example,
If n = 4 and k = 2, a solution is:

[  [2,4],  [3,4],  [2,3],  [1,2],  [1,3],  [1,4],]

/*第一个思路就是dfsn,k(XXX...XXX).size == kevery X = 1.2.3...n 用mark标记是否已经访问Submission Result: Time Limit Exceeded this method's output:combine(4, 2)12=13=14=21=23=24=31=32=34=41=42=43note:[1,2] == [2,1]按照题目的意思应该要标记起始点12=13=14=23=24=34*/public class Solution {public List<List<Integer>> list;public List<Integer> listItem;public boolean[] mark;public int nn;public int kk;    public List<List<Integer>> combine(int n, int k) {    list = new ArrayList<List<Integer>>();    listItem = new ArrayList<Integer>();    mark = new boolean[n];    kk = k;    nn = n;    getListItem(0);    return list;    }        private void getListItem(int step) {    if(step==kk) {    List<Integer> temp = new ArrayList<Integer>();    for(int i = 0;i<listItem.size();i++)    temp.add(listItem.get(i));    list.add(temp);    return;    }    for(int i = 0;i<nn;i++) {    if(!mark[i]) {    mark[i] = true;    if(listItem.size() == kk)    listItem.set(step, i+1);    else {    listItem.add(i+1);    }    getListItem(step+1);    mark[i] = false;    }    }    };}

/*对于这种固定长度解法step=0,有何种选择，选择的范围：起点终点，下一步的选择范围是什么直到step=k-1,共进行了k步，当step=k,也就是超出固定长度则处理returnSubmission Result: Accepted*/public class Solution {    public List<ArrayList<Integer>> combine(int n, int k) {    List<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();    if(k<=0 || n<1) return null;    getListItem(1,n,0,list,new int[k]);    return list;    }        private void getListItem(int start, int end, int step, List<ArrayList<Integer>> list, int[] path) {    if(step==path.length) {    ArrayList<Integer> listItem = new ArrayList<Integer>();    for(int i = 0;i<step;i++)    listItem.add(path[i]);    list.add(listItem);    return;    } else {    for(int i = start; i<= end;i++) {    path[step] = i;    getListItem(i+1, end, step+1, list, path);    }    }    };}

update 2015/04/14

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