博客园markdown太烂, 题解详情https://github.com/TangliziGit/leetcode/blob/master/solution/11-20.md
Leetcode Solution 11~20
marks:
@: hard to get a direct solution
%: need optimization
好题
%%% 11. Container With Most Water[Medium]
%%%%% 15. 3Sum[Medium]
%%% 16. 3Sum Closest [Medium]
% 18. 4Sum [Medium]
总结
- 初始化Stream: Stream.of(), Stream.iterate(0, x->x+1).limit(max)
- map: mapToInt(x -> ...), ...
- filter: filter(x -> ...)
- ending: reduce(Integer::min), findFirst()
- process: getAsInt(), orElse()
- 排序两边夹原理
设z=x+y;
if (z>tar) y<-; else x->;
作用：O(n)二元找x+yz相等（一元找xz相等用二分） - HashSet, HashMap 耗时严重 n次O(1)多花300ms
- 数组排序: Arrays.sort(nums)
- LinkedList方法: add
- 初始化List
: Arrays.asList(X, X, X, ...) - Stack用法: push(), peek(), pop(), isEmpty()
- int[] to List
:
Arrays.sort(arr);
Arrays.stream(arr).boxed().collect(Collectors.toList());
11. Container With Most Water[Medium]
%%% 11. Container With Most Water[Medium]
思路
- O(n^2)
- 树状数组从后向前区间更新最高高度（覆盖），然后从前遍历；再反向计算，取最大值 O(nlogn)
- 二叉搜索树 O(nlogn)
- 双指针,两边夹 O(n)
首先考虑一个解[i,j], 我们需要确定这个范围内解的最大值
在范围减小时, 要使解更大, 唯一的优势就是墙壁高度
所以每次更新时, 贪心的保护最高墙壁
暂时只能这样解释了...
要点
无
代码
class Solution {
public int maxArea(int[] height) {
int l=0, r=height.length-1, ans=0;
while (l<r){
int area=Math.min(height[l], height[r])*(r-l);
ans=Math.max(ans, area);
if (height[l]<height[r]) l++;
else r--;
}return ans;
}
}
12. Integer to Roman[Medium]
12. Integer to Roman[Medium]
思路
水题, 注意题意
要点
无
代码
class Solution {
private static String ans[]=new String[(int)4e3];
private static Integer[] value={
1, 4, 5, 9, 10,
40, 50, 90, 100,
400, 500, 900, 1000};
private static String[] expr={
"I", "IV", "V", "IX", "X",
"XL", "L", "XC", "C",
"CD", "D", "CM", "M"};
public String intToRoman(int num) {
return solve(num, value.length-1);
}
public String solve(int n, int ptr){
if (ptr==-1) return "";
if (ans[n]!=null) return ans[n];
ans[n]=repeat(expr[ptr], n/value[ptr])+
solve(n%value[ptr], ptr-1);
return ans[n];
}
public String repeat(String s, int n){
return new String(new char[n]).replace(" ", s);
}
}
13. Roman to Integer [Easy]
13. Roman to Integer [Easy]
思路
水题, 注意题意
要点
无
代码
class Solution {
private static Map<Character, Integer> map=new HashMap();
static{
map.put('I', 1);
map.put('V', 5);
map.put('X', 10);
map.put('L', 50);
map.put('C', 100);
map.put('D', 500);
map.put('M', 1000);
};
public int romanToInt(String s) {
int ans=0, len=s.length();
for (int i=0; i<len; i++)
if (i+1<len && map.get(s.charAt(i))<map.get(s.charAt(i+1)))
ans-=map.get(s.charAt(i));
else
ans+=map.get(s.charAt(i));
return ans;
}
}
14. Longest Common Prefix [Easy]
14. Longest Common Prefix [Easy]
思路
水题
刚好用来写Stream
要点
- 初始化Stream: Stream.of(), Stream.iterate(0, x->x+1).limit(max)
- map: mapToInt(x -> ...), ...
- filter: filter(x -> ...)
- ending: reduce(Integer::min), findFirst()
- process: getAsInt(), orElse()
代码
// Stream version
// 47ms, 38MB
class Solution {
public String longestCommonPrefix(String[] strs) {
if (strs==null || strs.length==0) return "";
int minlen=Stream.of(strs)
.mapToInt(x -> x.length())
.reduce(Integer::min)
.getAsInt();
int idx=Stream.iterate(0, x -> x+1).limit(0+minlen)
.filter(x -> check(strs, x))
.findFirst()
.orElse(minlen);
return strs[0].substring(0, idx);
}
public boolean check(String[] strs, int idx){
return Stream.of(strs)
.anyMatch(x -> x.charAt(idx)!=strs[0].charAt(idx));
}
}
// Original
// 4ms, 39MB
class OriginalSolution {
public String longestCommonPrefix(String[] strs) {
if (strs==null || strs.length==0) return "";
int idx=0, minlen=strs[0].length();
for (String str: strs)
minlen=Math.min(minlen, str.length());
for (;idx<minlen; idx++)
if (check(strs, idx)) break;
return strs[0].substring(0, idx);
}
public boolean check(String[] strs, int idx){
for (String str: strs)
if (str.charAt(idx)!=strs[0].charAt(idx))
return true;
return false;
}
}
15. 3Sum[Medium]
%%%%% 15. 3Sum[Medium]
思路
- O(n^3)
- O(n^2logn): for^2 + binarySearch
- O(n^2+nlogn+n) with a big constant:
取得两项和的map，然后遍历，最后去重 - O(n^2+nlogn+2n) with a small constant:
排序，得一个元素的对应下标map，若重复取最后
for^2 查找，插入HashSet - O(n^2) with a smaller constant:
for x: 两边夹找y+z==-x
要点
-
排序两边夹原理
设z=x+y;
if (z>tar) y<-; else x->;
作用：O(n)二元找x+yz相等（一元找xz相等用二分） -
HashSet, HashMap 耗时严重 n次O(1)多花300ms
代码
O(n^3) version
class Solution{
public List<List<Integer>> threeSum(int[] nums) {
LinkedList<List<Integer>> ans=new LinkedList();
Arrays.sort(nums);
for (int i=0; i<nums.length; i++)
if (i==0 || nums[i]!=nums[i-1]){
int l=i+1, r=nums.length-1;
while (l<r){
int sum=nums[l]+nums[r];
if (sum==-nums[i]){
ans.add(Arrays.asList(nums[i], nums[l], nums[r]));
while (l<r && nums[l]==nums[l+1]) l++;
while (l<r && nums[r]==nums[r-1]) r--;
}
if (sum>-nums[i]) r--;
else l++;
}
}
return ans;
}
}
O(n^2+nlogn+n) with a small constant
class Solution{
public List<List<Integer>> threeSum(int[] nums) {
Map<Integer, Integer> map=new HashMap();
Set<List<Integer>> set=new HashSet();
LinkedList<List<Integer>> ans=new LinkedList();
Arrays.sort(nums);
// if duplicated, use the last one
for (int i=0; i<nums.length; i++)
map.put(nums[i], i);
for (int i=0; i<nums.length; i++)
for (int j=i+1; j<nums.length; j++){
int sum=nums[i]+nums[j];
if (!map.containsKey(-sum)) continue;
if (map.get(-sum)<=j) continue;
set.add(Arrays.asList(nums[i], nums[j], nums[map.get(-sum)]));
}
for (List list: set)
ans.add(list);
return ans;
}
}
O(n^2+nlogn) solution
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
Map<Integer, List<Pair>> map=new HashMap();
Set<List<Integer>> set=new HashSet();
ArrayList<List<Integer>> ans=new ArrayList();
Arrays.sort(nums);
for (int i=0; i<nums.length; i++){
for (int j=i+1; j<nums.length; j++){
int sum=nums[i]+nums[j];
if (!map.containsKey(sum))
map.put(sum, new LinkedList<Pair>()); // O(1)
List<Pair> list=map.get(sum); // O(1)
list.add(new Pair(i, j)); // O(1)?
}
}
for (int i=0; i<nums.length; i++){
if (!map.containsKey(-nums[i])) continue;
for (Pair pair: map.get(-nums[i]))
if (i>pair.y && i!=pair.x && i!=pair.y){
set.add(Arrays.asList(nums[pair.x], nums[pair.y], nums[i]));
}
}
for (List<Integer> list: set)
ans.add(list);
return ans;
}
static class Pair{
final int x, y;
Pair(int x, int y){
this.x=x;
this.y=y;
}
}
}
16. 3Sum Closest [Medium]
%%% 16. 3Sum Closest [Medium]
思路
- O(n^3)
- O(n^2logn) 二分
- O(n^2) 双指针, 两边夹求最近, 因为这三道题都是双指针, 所以有点会用了
要点
- 数组排序: Arrays.sort(nums)
代码
class Solution {
public int threeSumClosest(int[] nums, int target) {
int ans=nums[0]+nums[1]+nums[2];
Arrays.sort(nums);
for (int i=0; i<nums.length; i++){
int l=i+1, r=nums.length-1;
while (l<r){
int sum=nums[i]+nums[l]+nums[r];
if (Math.abs(sum-target)<Math.abs(ans-target))
ans=sum;
if (sum<target) l++;
else r--;
}
}return ans;
}
}
17. Letter Combinations of a Phone Number [Medium]
17. Letter Combinations of a Phone Number [Medium]
思路
水题, 递归
要点
- LinkedList方法: add
- 初始化List
: Arrays.asList(X, X, X, ...)
代码
class Solution {
private String template="abcdefghijklmnopqrstuvwxyz";
public List<String> letterCombinations(String digits) {
if (digits.equals("")) return new ArrayList<String>();
return solve(digits, 0);
}
private List<String> solve(String digits, int ptr){
if (ptr==digits.length()) return Arrays.asList("");
List<String> tmp=solve(digits, ptr+1), ans=new LinkedList();
int num=digits.charAt(ptr)-'2', n=(num+2==9||num+2==7)?4:3;
int offset=(num+2==9||num+2==8)?1:0;
for (String str: tmp){
for (int i=offset; i<n+offset; i++)
ans.add(template.charAt(i+num*3)+str);
}return ans;
}
}
18. 4Sum [Medium]
% 18. 4Sum [Medium]
思路
- O(n^2+nlogn) with big constant:
map+set - O(n^3) with optimization:
for^2 双指针, 两边夹, 主要考虑一下双指针解法
要点
- int[] to List
:
Arrays.sort(arr);
Arrays.stream(arr).boxed().collect(Collectors.toList());
代码
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
int n=nums.length, size=0;
int[] pre=new int[n*(n-1)/2];
int[] pos=new int[n*(n-1)/2];
Map<Integer, List<Integer>> map=new HashMap();
Set<List<Integer>> set=new HashSet();
LinkedList<List<Integer>> ans=new LinkedList();
for (int i=0; i<n; i++)
for (int j=i+1; j<n; j++){
int sum=nums[i]+nums[j];
pre[size]=i;
pos[size++]=j;
if (!map.containsKey(sum))
map.put(sum, new LinkedList());
map.get(sum).add(size-1);
}
for (int i=0; i<n; i++)
for (int j=i+1; j<n; j++){
int rest=target-nums[i]-nums[j];
if (!map.containsKey(rest)) continue;
for (Integer ptr: map.get(rest))
if (i!=pre[ptr] && i!=pos[ptr] &&
j!=pre[ptr] && j!=pos[ptr]){
int[] l={nums[i], nums[j], nums[pre[ptr]], nums[pos[ptr]]};
Arrays.sort(l);
set.add(Arrays
.stream(l)
.boxed()
.collect(Collectors.toList()));
}
}
for (List<Integer> list: set)
ans.add(list);
return ans;
}
}
19. Remove Nth Node From End of List [Medium]
19. Remove Nth Node From End of List [Medium]
思路
水题
优化:
可以在第一个指针走了n个元素后, 在起一个指针, 等第一个结束了之后, 删除后一个指针的元素.
然而对复杂度没有提升, 而且有人说这是个很好的优化, 我说简直扯淡好吧
要点
无
代码
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
int len=0;
ListNode tmp=head;
while (tmp!=null){
len++;
tmp=tmp.next;
}
if (n==len) return head.next;
tmp=head;
for (int i=0; i<len-n-1; i++)
tmp=tmp.next;
tmp.next=tmp.next.next;
return head;
}
}
20. Valid Parentheses [Easy]
20. Valid Parentheses [Easy]
思路
水题, 栈
要点
- Stack用法: push(), peek(), pop(), isEmpty()
代码
class Solution {
private static Map<Character, Character> map=new HashMap();
static{
map.put('(', ')');
map.put('{', '}');
map.put('[', ']');
}
public boolean isValid(String s) {
Stack<Character> sta=new Stack();
int len=s.length();
for (int i=0; i<len; i++){
if (map.containsKey(s.charAt(i))) sta.push(s.charAt(i));
else{
if (!sta.isEmpty() && s.charAt(i)==map.get(sta.peek())) sta.pop();
else return false;
}
}
if (sta.isEmpty())
return true;
return false;
}
}