leetcode-Permutations
2015-01-09 10:05
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Given a collection of numbers, return all possible permutations.
For example,
and
public class Solution {
ArrayList<ArrayList<Integer>> result=new ArrayList<ArrayList<Integer>>();
public ArrayList<ArrayList<Integer>> permute(int[] num) {
if(num.length==0)return null;
myPermute(num,0,new ArrayList<Integer>());
return result;
}
public void myPermute(int[] num,int index,ArrayList<Integer> cur){
for(int i=0;i<=index;i++){
cur.add(i,num[index]);
ArrayList<Integer> a=new ArrayList(cur);
if(index==num.length-1)result.add(a); // we should create a new list because the cur list will
For example,
[1,2,3]have the following permutations:
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
and
[3,2,1].
public class Solution {
ArrayList<ArrayList<Integer>> result=new ArrayList<ArrayList<Integer>>();
public ArrayList<ArrayList<Integer>> permute(int[] num) {
if(num.length==0)return null;
myPermute(num,0,new ArrayList<Integer>());
return result;
}
public void myPermute(int[] num,int index,ArrayList<Integer> cur){
for(int i=0;i<=index;i++){
cur.add(i,num[index]);
ArrayList<Integer> a=new ArrayList(cur);
if(index==num.length-1)result.add(a); // we should create a new list because the cur list will
need to be modified later else if(index<num.length-1)myPermute(num,index+1,cur); cur.remove(i); //undo effect } } }
这种是backtracking 的典型题目:
思路都是一样的,在每一步思考在这一步有哪些情况,然后分别尝试,在每一次尝试后都要undo 这一步的effect,在符合条件的时候把结果村粗在相应的地方。(我的习惯是在函数中输入相应的参数传递下来)
然后技巧和减小时间复杂度的方法就是有些情况我们是可以排除的,这也是最难的部分。
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