LeetCode数组习题
26.Remove Duplicates from Sorted Array
题目描述:
Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.
Do not allocate extra space for another array, you must do this in place with constant mem- ory.
For example, Given input array A = [1,1,2],
Your function should return length = 2, and A is now [1,2].
中文:
一个有序数组,返回不重复的元素的个数。
Java代码
public int removeDuplicates(int[] nums) {
int index=0;
for(int i=1;i<nums.length;++i){
if (nums[i]!=nums[index]){
index=index+1;
nums[index]=nums[i];
}
}
return index+1;
}
时间复杂度 | 空间复杂度 |
---|---|
O(n) | O(1) |
80. Remove Duplicates from Sorted Array II
Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
For example,
Given sorted array nums = [1,1,1,2,2,3],
Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn't matter what you leave beyond the new length.
Java代码
public int removeDuplicates(int[] nums) {
int index=0;
int occur=1;
for(int i=1;i<nums.length;i++){
if(nums[index]==nums[i]){
if(occur<2){
index=index+1;
nums[index]=nums[i];
occur++;
}
}else{
index=index+1;
nums[index]=nums[i];
occur=1;
}
}
return index+1;
}
时间复杂度 | 空间复杂度 |
---|---|
O(n) | O(1) |
1. Two Sum
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
中文:
给定一个整型数组和一个目标值,返回和为目标值的2个数的位置。
解法一:冒泡排序思想遍历
public int[] twoSum(int[] nums, int target) {
int[] index = new int[2];
for (int i = 0; i < nums.length-1; i++) {
for (int j=i+1;j<nums.length;j++){
if (nums[i] +nums[j]==target){
index[0]=i;
index[1]=j;
break;
}
}
}
return index;
}
时间复杂度 | 空间复杂度 |
---|---|
O(n^2) | O(1) |
解法二:使用HashMap
public int[] twoSum(int[] nums, int target) {
int[] index = new int[2];
HashMap<Integer,Integer> map=new HashMap<>();
for (int i = 0; i < nums.length; i++) {
if (map.containsKey(target-nums[i])){
index[0]=map.get(target-nums[i]);
index[1]=i;
}
map.put(nums[i],i);
}
return index;
}
时间复杂度 | 空间复杂度 |
---|---|
O(n) | O(n) |
7. Reverse Integer
题目
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
click to show spoilers.
Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
中文:
整数逆序输出,超过32位无符号数的返回0.
Java代码:
public int inverse(int x) {
long result = 0;
while (x != 0) {
result = result * 10 + x % 10;
if (result > Integer.MAX_VALUE || result < Integer.MIN_VALUE) {
return 0;
}
x = x / 10;
}
return (int) result;
}
时间复杂度 | 空间复杂度 |
---|---|
O(n) | O(1) |
4. Median of Two Sorted Arrays
有序数组nums1和nums2长度分别为m和n,返回中位数。要求时间复杂度为O(log(m+n))。
例子1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
例子2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
先合并有序数组,再返回中位数,java代码如下:
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int m=nums1.length;
int n=nums2.length;
int N=m+n;
int[] nums=new int[N];
int i=0,j=0,k=0;
while (i<m&&j<n){
if (nums1[i]<nums2[j]){
nums[k++]=nums1[i++];
}else{
nums[k++]=nums2[j++];
}
}
while (i<m) nums[k++]=nums1[i++];
while (j<n) nums[k++]=nums2[j++];
double result=0;
if (N%2!=0){
result=(double) nums[N/2];
}else{
result =(double)(nums[N/2-1]+nums[N/2])/2;
}
return result;
}
时间复杂度 | 空间复杂度 |
---|---|
O(log(m+n)) | O(log(m+n)) |
136. Single Number
给定一个整型数组,数组里的元素都出现2次,只有1个除外。找出只出现1次的元素:
解法一
Java代码:
public int singleNumber(int[] nums) {
Map<Integer,Integer> map=new HashMap<>();
for (int a:nums){
if (map.containsKey(a)){
map.remove(a);
}else{
map.put(a,1);
}
}
return (int)map.keySet().toArray()[0];
}
解法二
通过求与操作:
public int singleNumber(int[] nums) {
int n = nums.length;
int goal=nums[0];
for (int i=1;i<n;i++){
goal^=nums[i];
}
return goal;
}
189. Rotate Array
For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].
思路:三步反转法
java代码:
class Solution {
public void rotate(int[] nums, int k) {
k%=nums.length;
reverseArr(nums,0,nums.length-1);
reverseArr(nums,0,k-1);
reverseArr(nums,k,nums.length-1);
}
public void reverseArr(int[] nums,int from,int to){
while(from<to){
int temp=nums[from];
nums[from]=nums[to];
nums[to]=temp;
from++;
to--;
}
}
}
时间复杂度 | 空间复杂度 |
---|---|
O(n) | O(1) |
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