[Leetcode] solution to a problem record 11-20

Park markdown blog sucks, problem solution details https://github.com/TangliziGit/leetcode/blob/master/solution/11-20.md

Leetcode Solution 11~20

marks:
@: hard to get a direct solution
%: need optimization

Good question

%%% 11. Container With Most Water[Medium]
%%%%% 15. 3Sum[Medium]
%%% 16. 3Sum Closest [Medium]
% 18. 4Sum [Medium]

to sum up

1. Initialization Stream: Stream.of (), Stream.iterate (0, x-> x + 1) .limit (max)
2. map: mapToInt(x -> ...), ...
3. filter: filter(x -> ...)
4. ending: reduce(Integer::min), findFirst()
5. process: getAsInt(), orElse()
6. Sorting principles interposed sides
   provided X + Y = Z;
   IF (Z> the tar) Y <-; X-the else>;
   effect: O (n) to find the two yuan equal to x + y == z (mono equal to find x == z by half)
   
7. HashSet, HashMap takes serious n times O (1) spend 300ms
8. Array Sort: Arrays.sort (nums)
9. LinkedList methods: add
10. Initialization List : Arrays.asList(X, X, X, ...)
11. Stack Usage: push (), peek (), pop (), isEmpty ()
12. 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]

Thinking

1. O (n ^ 2)
2. Fenwick tree forwardly from the rear section to update the maximum height (cover), and then used to traverse; then reverse calculation, taking the maximum O (nlogn)
3. Binary search tree O (nlogn)
4. Double pointer, clamp both sides of O (n)
   Consider first a solution of [i, j], we need to determine the maximum value of the solution within this range
   when the range is reduced to make the solution more, the only advantage is the height of the wall
   so that each update when, greedy highest protection wall
   being only explains the ...

Points

no

Code

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 ```java 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("\0", s); } } ```

13. Roman to Integer [Easy]

13. Roman to Integer [Easy]

Thinking

Mizu题, attention 题意

Points

no

Code

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]

Thinking

Water problems
just to write Stream

Points

1. Initialization Stream: Stream.of (), Stream.iterate (0, x-> x + 1) .limit (max)
2. map: mapToInt(x -> ...), ...
3. filter: filter(x -> ...)
4. ending: reduce(Integer::min), findFirst()
5. process: getAsInt(), orElse()

Code

// 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] ### ideas 1. O (n ^ 3) 2. O (n ^ 2logn): for ^ 2 + binarySearch 3. O (n ^ 2 + nlogn + n ) with a big constant: and the two acquired map, then traverse, and finally to the weight 4. O (n ^ 2 + nlogn + 2n) with a small constant: sorting, have a map element corresponding to the index, if repeated taking Finally, for ^ 2 search, insert HashSet 5. O (n ^ 2) with a smaller constant: for x: sandwiched on both sides to find y + z == - x ### points 1 ** ** sort disposed on both sides of the principles interposed z = x + y; if (z> tar) y <-; else x->; effect: O (n) to find the two yuan equal to x + y == z (mono find x == z equal a half) 2. HashSet , HashMap takes serious n times O (1) spend 300ms ### Code O (n ^ 3) version `` `java class Solution {public List > threeSum(int[] nums) { LinkedList > ans=new LinkedList(); Arrays.sort(nums); for (int i=0; i -nums[i]) r--; else l++; } } return ans; } } ``` O(n^2+nlogn+n) with a small constant ```java class Solution{ public List > threeSum(int[] nums) { Map map=new HashMap(); Set > set=new HashSet(); LinkedList > ans=new LinkedList(); Arrays.sort(nums); // if duplicated, use the last one for (int i=0; i > threeSum(int[] nums) { Map > map=new HashMap(); Set > set=new HashSet(); ArrayList > ans=new ArrayList(); Arrays.sort(nums); for (int i=0; i ()); // O(1) List list=map.get(sum); // O(1) list.add(new Pair(i, j)); // O(1)? } } for (int i=0; i pair.y && i!=pair.x && i!=pair.y){ set.add(Arrays.asList(nums[pair.x], nums[pair.y], nums[i])); } } for (List 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]

Thinking

1. O (n ^ 3)
2. O (n ^ 2logn) bipartite
3. O (n ^ 2) the double pointer, both sides of the clip seeking Recently, because these three questions are double pointer, the bit will be used

Points

1. Array Sort: Arrays.sort (nums)

Code

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]

Thinking

Water problems, recursive

Points

1. LinkedList methods: add
2. Initialization List : Arrays.asList(X, X, X, ...)

Code

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] ### ideas 1. O (n ^ 2 + nlogn) with big constant: map + set 2. O (n ^ 3) with optimization: for ^ 2 double pointer, both sides of the clip, The main solution to think about double pointer ### points 1. int [] to List : ```java Arrays.sort(arr); Arrays.stream(arr).boxed().collect(Collectors.toList()); ``` ### 代码 ```java class Solution { public List > 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 > map=new HashMap(); Set > set=new HashSet(); LinkedList > ans=new LinkedList(); for (int i=0; i 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]

Thinking

Water problems
Optimization:
can go in the first n elements of a pointer later, from a pointer, after the end of a ranking, after deleting a pointer element.
However, the degree of complexity does not improve, but some people say this is a very good optimization, I say simply nonsense well

Points

no

Code

/**
 * 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]

Thinking

Water problems, stack

Points

1. Stack Usage: push (), peek (), pop (), isEmpty ()

Code

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;
    }
}