1. Permutations
Given a collection of distinct numbers, return all possible permutations.
For example,[1,2,3] have the following permutations:
[ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ]
思路；直接使用递归来遍历即可，因为数字是不重复的所以，所以这里的每层递归都从索引0开始遍历整个数组，将不在集合中的数字添加到集合中，到集合元素个数达到要求时返回到上层递归即可。
public List<List<Integer>> permute(int[] nums) { List<List<Integer>> list = new ArrayList<>(); backtrack(list, new ArrayList<>(), nums); return list; } private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums){ if(tempList.size() == nums.length){ list.add(new ArrayList<>(tempList)); } else{ for(int i = 0; i < nums.length; i++){ if(tempList.contains(nums[i])) continue; // element already exists, skip tempList.add(nums[i]); backtrack(list, tempList, nums); tempList.remove(tempList.size() - 1); //注意这里remove方法是按照索引来删的，因为参数是int类型 } } }
2. Permutations II
Given a collection of numbers that might contain duplicates, return all possible unique permutations
For example,[1,1,2] have the following unique permutations:
[ [1,1,2], [1,2,1], [2,1,1] ]
思路：和上一题不同在于数组中可能含有重复数字。解法和上题类似，但是递归遍历数组时要考虑到的是因为数组中有重复数字，所以遇到相同数时要做判断，一是如果是用过的数字则不能再用，利用一个布尔数组来记录，还有一个避免对连着的重复数字循环遍历，以 1，1，2和1，1，1来举例，找所有的组合仍然是一层层从索引0到2循环遍历，总共有3层，所以对于 1，1来说第一层遍历到第二个1，第二层遍历到第一个1和第一层遍历到第一个1，第二层遍历到第二个1这两种情况是完全相同的，因此需要避免这种重复的情况，因此对元数组排序一遍后，对连着的重复数字只需要递归遍历一遍即可，如果当前数字和前面数字相同并且前面数字还没用过的情形则直接跳过。
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> list = new ArrayList<>();
Arrays.sort(nums);
backtrack(list, new ArrayList<>(), nums, new boolean[nums.length]);
return list;
}
private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, boolean [] used){
if(tempList.size() == nums.length){
list.add(new ArrayList<>(tempList));
} else{
for(int i = 0; i < nums.length; i++){
// 当nums[i]用过了或者nums[i]和前一个数字相等且前一个数字没用过的情形则直接跳过
if(used[i] || (i > 0 && nums[i] == nums[i-1] && !used[i - 1])) continue;
used[i] = true;
tempList.add(nums[i]);
backtrack(list, tempList, nums, used);
used[i] = false; //用完后记得置回为false
tempList.remove(tempList.size() - 1);
}
}
}
3. Rotate Image
You are given an n x n 2D matrix representing an image.
Rotate the image by 90 degrees (clockwise).
Note:
You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.
Example 1:
Given input matrix = [ [1,2,3], [4,5,6], [7,8,9] ], rotate the input matrix in-place such that it becomes: [ [7,4,1], [8,5,2], [9,6,3] ]
Example 2:
Given input matrix = [ [ 5, 1, 9,11], [ 2, 4, 8,10], [13, 3, 6, 7], [15,14,12,16] ], rotate the input matrix in-place such that it becomes: [ [15,13, 2, 5], [14, 3, 4, 1], [12, 6, 8, 9], [16, 7,10,11] ]
思路：本来想根据索引替换规律来解，但是以此遍历整个数组时会出现重复替换的问题。因为顺时针旋转矩阵其实就是将行变成列，再变动列号。所以比较简单的做法是先对矩阵进行转置，然后按列交换矩阵：
The idea was firstly transpose the matrix and then flip it symmetrically. For instance,
1 2 3
4 5 6
7 8 9
after transpose, it will be swap(matrix[i][j], matrix[j][i])
1 4 7
2 5 8
3 6 9
Then flip the matrix horizontally. (swap(matrix[i][j], matrix[i][matrix.length-1-j])
7 4 1
8 5 2
9 6 3写代码时一个比较麻烦的地方是怎么输入一个矩阵
import java.util.*;
public class LeetCode{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int n=Integer.parseInt(sc.nextLine());
int[][] matrix=new int[n][n];
for(int i=0;i<n;i++){
String str=sc.nextLine();
String[] strs=str.split(" |,");
for(int j=0;j<n;j++){
matrix[i][j]=Integer.parseInt(strs[j]);
}
}
rotate(matrix);
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
System.out.print(matrix[i][j]+" ");
}
System.out.println();
}
}
static void rotate(int[][] matrix) {
for(int i = 0; i<matrix.length; i++){
for(int j = i; j<matrix[0].length; j++){
int temp = 0;
temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
for(int i =0 ; i<matrix.length; i++){
for(int j = 0; j<matrix.length/2; j++){
int temp = 0;
temp = matrix[i][j];
matrix[i][j] = matrix[i][matrix.length-1-j];
matrix[i][matrix.length-1-j] = temp;
}
}
}
}
