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Commit99acaf0

Browse files
author
applewjg
committed
permutations
Change-Id: I4b4cac41ef886ca1bc53090fa500f433873d5262
1 parent2e1678b commit99acaf0

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2 files changed

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-0
lines changed

2 files changed

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‎Permutations.h

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/*
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Author: King, wangjingui@outlook.com
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Date: Dec 25, 2014
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Problem: Permutations
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Difficulty: Easy
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Source: https://oj.leetcode.com/problems/permutations/
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Notes:
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Given a collection of numbers, return all possible permutations.
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For example,
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[1,2,3] have the following permutations:
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[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].
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Solution: dfs...
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*/
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publicclassSolution {
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publicList<List<Integer>>permute_1(int[]num) {
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List<List<Integer>>res=newArrayList<List<Integer>>();
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List<Integer>path=newArrayList<Integer>();
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boolean[]free=newboolean[num.length];
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Arrays.fill(free, true);
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permuteRe(num,res,path,free);
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returnres;
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}
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voidpermuteRe(int[]num,List<List<Integer>>res,List<Integer>path,boolean[]free) {
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if(path.size()==num.length) {
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ArrayList<Integer>tmp=newArrayList<Integer>(path);
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res.add(tmp);
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return;
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}
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for (inti=0;i<num.length;++i) {
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if (free[i]== true) {
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free[i]= false;
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path.add(num[i]);
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permuteRe(num,res,path,free);
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path.remove(path.size()-1);
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free[i]= true;
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}
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}
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}
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publicbooleannextPermutation(int[]num) {
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intlast=num.length-1;
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inti=last;
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while (i>0&&num[i-1] >=num [i])--i;
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for (intl=i,r=last;l<r;++l,--r) {
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num[l]=num[l] ^num[r];
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num[r]=num[l] ^num[r];
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num[l]=num[l] ^num[r];
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}
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if (i==0) {
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return false;
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}
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intj=i;
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while (j <=last&&num[i-1] >=num[j])++j;
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num[i-1]=num[i-1] ^num[j];
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num[j]=num[i-1] ^num[j];
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num[i-1]=num[i-1] ^num[j];
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return true;
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}
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publicList<List<Integer>>permute_2(int[]num) {
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List<List<Integer>>res=newArrayList<List<Integer>>();
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Arrays.sort(num);
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do {
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List<Integer>path=newArrayList<Integer>();
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for (inti :num)path.add(i);
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res.add(path);
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}while(nextPermutation(num));
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returnres;
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}
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}

‎PermutationsII.h

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/*
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Author: King, wangjingui@outlook.com
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Date: Dec 25, 2014
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Problem: Permutations II
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Difficulty: Easy
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Source: https://oj.leetcode.com/problems/permutations-ii/
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Notes:
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Given a collection of numbers that might contain duplicates, return all possible unique permutations.
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For example,
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[1,1,2] have the following unique permutations:
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[1,1,2], [1,2,1], and [2,1,1].
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Solution: dfs...
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*/
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publicclassSolution {
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publicList<List<Integer>>permuteUnique_1(int[]num) {
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List<List<Integer>>res=newArrayList<List<Integer>>();
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List<Integer>path=newArrayList<Integer>();
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Arrays.sort(num);
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boolean[]visited=newboolean[num.length];
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Arrays.fill(visited, false);
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permuteRe(num,res,path,visited);
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returnres;
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}
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voidpermuteRe(int[]num,List<List<Integer>>res,List<Integer>path,boolean[]visited) {
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if(path.size()==num.length) {
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ArrayList<Integer>tmp=newArrayList<Integer>(path);
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res.add(tmp);
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return;
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}
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for (inti=0;i<num.length;++i) {
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if(visited[i]||(i!=0&&num[i-1]==num[i]&&visited[i-1]))continue;
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visited[i]= true;
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path.add(num[i]);
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permuteRe(num,res,path,visited);
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path.remove(path.size()-1);
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visited[i]= false;
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}
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}
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publicbooleannextPermutation(int[]num) {
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intlast=num.length-1;
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inti=last;
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while (i>0&&num[i-1] >=num [i])--i;
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for (intl=i,r=last;l<r;++l,--r) {
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num[l]=num[l] ^num[r];
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num[r]=num[l] ^num[r];
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num[l]=num[l] ^num[r];
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}
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if (i==0) {
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return false;
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}
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intj=i;
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while (j <=last&&num[i-1] >=num[j])++j;
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num[i-1]=num[i-1] ^num[j];
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num[j]=num[i-1] ^num[j];
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num[i-1]=num[i-1] ^num[j];
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return true;
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}
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publicList<List<Integer>>permuteUnique_2(int[]num) {
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List<List<Integer>>res=newArrayList<List<Integer>>();
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Arrays.sort(num);
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do {
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List<Integer>path=newArrayList<Integer>();
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for (inti :num)path.add(i);
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res.add(path);
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}while(nextPermutation(num));
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returnres;
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}
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}

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