491.递增子序列 leetcode.com/problems/no…
class Solution {
List<List<Integer>> result = new ArrayList<>();
List<Integer> path = new ArrayList<>();
public List<List<Integer>> findSubsequences(int[] nums) {
backTracking(nums, 0);
return result;
}
private void backTracking(int[] nums, int startIndex){
if(path.size() >= 2)
result.add(new ArrayList<>(path));
HashSet<Integer> hs = new HashSet<>();
for(int i = startIndex; i < nums.length; i++){
if(!path.isEmpty() && path.get(path.size() -1 ) > nums[i] || hs.contains(nums[i]))
continue;
hs.add(nums[i]);
path.add(nums[i]);
backTracking(nums, i + 1);
path.remove(path.size() - 1);
}
}
}
46.全排列 leetcode.com/problems/pe…
class Solution {
List<List<Integer>> result = new ArrayList<>();// 存放符合条件结果的集合
LinkedList<Integer> path = new LinkedList<>();// 用来存放符合条件结果
boolean[] used;
public List<List<Integer>> permute(int[] nums) {
if (nums.length == 0){
return result;
}
used = new boolean[nums.length];
permuteHelper(nums);
return result;
}
private void permuteHelper(int[] nums){
if (path.size() == nums.length){
result.add(new ArrayList<>(path));
return;
}
for (int i = 0; i < nums.length; i++){
if (used[i]){
continue;
}
used[i] = true;
path.add(nums[i]);
permuteHelper(nums);
path.removeLast();
used[i] = false;
}
}
}
47.全排列 II leetcode.com/problems/pe…
class Solution {
//存放结果
List<List<Integer>> result = new ArrayList<>();
//暂存结果
List<Integer> path = new ArrayList<>();
public List<List<Integer>> permuteUnique(int[] nums) {
boolean[] used = new boolean[nums.length];
Arrays.fill(used, false);
Arrays.sort(nums);
backTrack(nums, used);
return result;
}
private void backTrack(int[] nums, boolean[] used) {
if (path.size() == nums.length) {
result.add(new ArrayList<>(path));
return;
}
for (int i = 0; i < nums.length; i++) {
// used[i - 1] == true,说明同⼀树⽀nums[i - 1]使⽤过
// used[i - 1] == false,说明同⼀树层nums[i - 1]使⽤过
// 如果同⼀树层nums[i - 1]使⽤过则直接跳过
if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) {
continue;
}
//如果同⼀树⽀nums[i]没使⽤过开始处理
if (used[i] == false) {
used[i] = true;//标记同⼀树⽀nums[i]使⽤过,防止同一树枝重复使用
path.add(nums[i]);
backTrack(nums, used);
path.remove(path.size() - 1);//回溯,说明同⼀树层nums[i]使⽤过,防止下一树层重复
used[i] = false;//回溯
}
}
}
}
332.重新安排行程 leetcode.com/problems/re…
class Solution {
private LinkedList<String> res;
private LinkedList<String> path = new LinkedList<>();
public List<String> findItinerary(List<List<String>> tickets) {
Collections.sort(tickets, (a, b) -> a.get(1).compareTo(b.get(1)));
path.add("JFK");
boolean[] used = new boolean[tickets.size()];
backTracking((ArrayList) tickets, used);
return res;
}
public boolean backTracking(ArrayList<List<String>> tickets, boolean[] used) {
if (path.size() == tickets.size() + 1) {
res = new LinkedList(path);
return true;
}
for (int i = 0; i < tickets.size(); i++) {
if (!used[i] && tickets.get(i).get(0).equals(path.getLast())) {
path.add(tickets.get(i).get(1));
used[i] = true;
if (backTracking(tickets, used)) {
return true;
}
used[i] = false;
path.removeLast();
}
}
return false;
}
}
class Solution {
List<List<String>> res = new ArrayList<>();
public List<List<String>> solveNQueens(int n) {
char[][] chessboard = new char[n][n];
for (char[] c : chessboard) {
Arrays.fill(c, '.');
}
backTrack(n, 0, chessboard);
return res;
}
public void backTrack(int n, int row, char[][] chessboard) {
if (row == n) {
res.add(Array2List(chessboard));
return;
}
for (int col = 0;col < n; ++col) {
if (isValid (row, col, n, chessboard)) {
chessboard[row][col] = 'Q';
backTrack(n, row+1, chessboard);
chessboard[row][col] = '.';
}
}
}
public List Array2List(char[][] chessboard) {
List<String> list = new ArrayList<>();
for (char[] c : chessboard) {
list.add(String.copyValueOf(c));
}
return list;
}
public boolean isValid(int row, int col, int n, char[][] chessboard) {
// 检查列
for (int i=0; i<row; ++i) { // 相当于剪枝
if (chessboard[i][col] == 'Q') {
return false;
}
}
// 检查45度对角线
for (int i=row-1, j=col-1; i>=0 && j>=0; i--, j--) {
if (chessboard[i][j] == 'Q') {
return false;
}
}
// 检查135度对角线
for (int i=row-1, j=col+1; i>=0 && j<=n-1; i--, j++) {
if (chessboard[i][j] == 'Q') {
return false;
}
}
return true;
}
}
TODO