1. Rotate List
Given a list, rotate the list to the right by k places, where k is non-negative.
Example:
Given 1->2->3->4->5->NULL and k = 2, return 4->5->1->2->3->NULL.
思路：这题感觉有点没有说清楚，如果k大于了链表长度该怎么办呢？这个时候怎么确定旋转位置呢？从右往左数，超出数组长度则取余。所以第一步是确定数组长度，结合k来确定旋转位置，然后再作旋转。
Since n may be a large number compared to the length of list. So we need to know the length of linked list.After that, move the list after the (l-n%l )th node to the front to finish the rotation.
Ex: {1,2,3} k=2 Move the list after the 1st node to the front
Ex: {1,2,3} k=5, In this case Move the list after (3-5%3=1)st node to the front.
public ListNode rotateRight(ListNode head, int n) {
if (head==null||head.next==null) return head;
ListNode dummy=new ListNode(0);
dummy.next=head;
ListNode fast=dummy,slow=dummy;
int i;
for (i=0;fast.next!=null;i++)//Get the total length
fast=fast.next;
for (int j=i-n%i;j>0;j--) //Get the i-n%i th node
slow=slow.next;
fast.next=dummy.next; //Do the rotation
dummy.next=slow.next;
slow.next=null;
return dummy.next;
}
2. Set Matrix Zeroes
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
思路：这题可以用遍历的方式取求解，但关键是如何在O(1)的空间复杂度内求解这题。一个可行的想法是，先遍历一遍矩阵，将元素为0的那一行和那一列的第一个元素标记为0，然后从第二行第二列开始遍历矩阵。如果这行或者这列的开始元素是0则将这个元素置为0，剩下一个要做的是处理将第一行和第一列上要置为0的元素。
import java.util.*; public class LeetCode{ public static void main(String[] args){ Scanner sc=new Scanner(System.in); int m=sc.nextInt(); int n=sc.nextInt(); int[][] matrix=new int[m][n]; // 老实说输入一个矩阵真的是超级烦，这里不加这个就会输入异常，大概是因为上面的nextInt()不会读入换行符？ sc.nextLine(); for(int i=0;i<m;i++){ String input=sc.nextLine(); String[] inputs=input.split(","); for(int j=0;j<inputs.length;j++){ matrix[i][j]=Integer.parseInt(inputs[j]); } } SetZeros(matrix); for(int i=0;i<matrix.length;i++){ for(int j=0;j<matrix.length;j++){ System.out.print(matrix[i][j]+" "); } System.out.println(); } } static void SetZeros(int[][] matrix){ boolean fr=false, fc=false; for(int i=0;i<matrix.length;i++){ for(int j=0;j<matrix[0].length;j++){ if(matrix[i][j]==0){ if(i==0)fr=true; if(j==0)fc=true; matrix[i][0]=0; matrix[0][j]=0; } } } for(int i=1;i<matrix.length;i++){ for(int j=1;j<matrix.length;j++){ if(matrix[i][0]==0||matrix[0][j]==0) matrix[i][j]=0; } } if(fr==true){ for(int j=0;j<matrix[0].length;j++) matrix[0][j]=0; } if(fc==true){ for(int i=0;i<matrix.length;i++) matrix[i][0]=0; } } }
3. Sort Colors
Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue.
Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.
Note:
You are not suppose to use the library's sort function for this problem.
Follow up:
A rather straight forward solution is a two-pass algorithm using counting sort.
First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total number of 0's, then 1's and followed by 2's.
Could you come up with an one-pass algorithm using only constant space?
import java.util.*;
public class LeetCode{
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
String input=sc.nextLine();
String[] inputs=input.split(",");
int[] colors=new int[inputs.length];
for(int i=0;i<inputs.length;i++){
colors[i]=Integer.parseInt(inputs[i]);
}
sortColors(colors);
for(int c:colors){
System.out.print(c+" ");
}
}
static void sortColors(int[] colors){
if(colors==null||colors.length<2)
return;
int low=0,high=colors.length-1;
for(int i=low;i<=high;){
if(colors[i]==0){
int temp=colors[i];
colors[i]=colors[low];
colors[low]=temp;
i++;low++;
}else if(colors[i]==2){
int temp=colors[i];
colors[i]=colors[high];
colors[high]=temp;
high--;
}else{
i++;
}
}
}
}
