Day 4 of leetcode question brushing

1. A string that concatenates all words

 Given a string  s  And some of the   Same length   's words  words . find  s  It happens that  words  The starting position of the substring formed by concatenation of all words in . Pay attention to the relationship between the string and  words  The words in match exactly , No other characters in the middle  , But there's no need to think about  words  The order in which words are concatenated . 
 Example  1：  
 Input ：s = "barfoothefoobarman", words = ["foo","bar"]
 Output ：[0,9]
 explain ：
 From the index  0  and  9  The first substrings are  "barfoo"  and  "foobar" .
 The order of output is not important , [9,0]  It's also a valid answer . 
 Example  2：  
 Input ：s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
 Output ：[]
 Example  3：  
 Input ：s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]
 Output ：[6,9,12]

analysis ：

Using the slide window . Because the word length is fixed , So we can calculate whether the number of words intercepted from the string is equal to  words  Li equivalency , So we can borrow hash table . One is hash table  words, A hash table is a truncated string , Compare whether two hashes are equal .

public static List<Integer> solution2(String s, String[] words) {
        List<Integer> res = new ArrayList<>();
        Map<String, Integer> wordsMap = new HashMap<>();
        if (s.length() == 0 || words.length == 0) return res;
        for (String word: words) {
            //  Main string s There is no such word in English , Direct return null 
            if (s.indexOf(word) < 0) return res;
            // map Save each word in , And the number of times it appears 
            wordsMap.put(word, wordsMap.getOrDefault(word, 0) + 1);
        }
        //  The length of each word ,  Total length 
        int oneLen = words[0].length(), wordsLen = oneLen * words.length;
        //  Main string s The length is less than the sum of words , Returns an empty 
        if (wordsLen > s.length()) return res;
        //  Only discuss from 0,1,..., oneLen-1  The starting substring ,
        //  The window size for each match is  wordsLen, Move back one word at a time , The current window position is maintained by the left and right windows 
        for (int i = 0; i < oneLen; ++i) {
            //  Left and right windows 
            int left = i, right = i, count = 0;
            //  Count each qualified word
            Map<String, Integer> subMap = new HashMap<>();
            //  The right window cannot exceed the length of the main string 
            while (right + oneLen <= s.length()) {
                //  Get a word 
                String word = s.substring(right, right + oneLen);
                //  There is a window moving right 
                right += oneLen;
                // words[] There is no such word in English , Then the current window must fail to match , Move directly to the right after this word 
                if (!wordsMap.containsKey(word)) {
                    left = right;
                    //  Word statistics in the window map Empty , Re count 
                    subMap.clear();
                    //  Count the words that meet the requirements 0
                    count = 0;
                } else {
                    //  Count the number of occurrences of this word in the current substring 
                    subMap.put(word, subMap.getOrDefault(word, 0) + 1);
                    ++count;
                    //  If this word appears more than words[] The number of times it corresponds to , And because of every match and words Substrings of equal length 
                    //  Such as  ["foo","bar","foo","the"]  "| foobarfoobar| foothe"
                    //  the second bar Although it is words[] The words in , But the number is copied , Then shift the length of one word to the right  "|barfoobarfoo|the"
                    // bar Still not in line with , So directly from this inconformity bar Then start matching , That is to say, the non-conforming bar And the words before it ( strand ) Move it all out 
                    while (subMap.getOrDefault(word, 0) > wordsMap.getOrDefault(word, 0)) {
                        //  Count characters from the current window map Delete all words from the left window to the number limit ( Times minus one )
                        String w = s.substring(left, left + oneLen);
                        subMap.put(w, subMap.getOrDefault(w, 0) - 1);
                        //  Subtract one from the number of words that match 
                        --count;
                        //  Move the left window position to the right 
                        left += oneLen;
                    }
                    //  The current window string meets the requirements 
                    if (count == words.length) res.add(left);
                }
            }
        }
        return res;
    }

2. Binary 1 The number of

Write a function , Input is an unsigned integer （ In the form of a binary string ）, Returns the number of digits in its binary expression as '1' The number of ）.

Input ：n = 11 ( Console input 00000000000000000000000000001011)
Output ：3
explain ： Binary string of input 00000000000000000000000000001011  in , There are three of them '1'.

Input ：n = 128 ( Console input 00000000000000000000000010000000)
Output ：1
explain ： Binary string of input 00000000000000000000000010000000  in , There is one in all for '1'.

analysis ：

And operation &： There is a binary number n, if  n&1 = 0 , be  n  The rightmost bit of binary is  0; if n&1 = 1, Then the rightmost bit of binary is 1.

Algorithm flow ：
Initialize the quantity statistics variable res = 0.
Loop bit by bit ： When n = 0 Jump out when .
res += n & 1 ： if n \& 1 = 1, Then statistics res  Add one .
n >>= 1 ： Put the binary number n  Move one unsigned right （ Java Right shift without sign in to ">>>>>>" ） .
Return statistics quantity res .

public class Solution {
    public int hammingWeight(int n) {
        int res = 0;
        while(n != 0) {
            res += n & 1;
            n >>>= 1;
        }
        return res;
    }
}

3. Integer power of value

Realization pow(x,n), Computation x Of n Power function . Do not use library functions , At the same time, there is no need to consider the problem of large numbers .

analysis ：

class Solution {
    public double myPow(double x, int n) {
        return n>0?mult(x,n):1/mult(x,-n);
    }

    public double mult(double x, int n){
        if(n==0)
            return 1.0;
        double y = mult(x, n/2);
        return n%2==0?y*y:y*y*x;
    }
}

4.  Input number  n, Print out from... In order 1 To the biggest n A decimal number . Such as input 3, Then print out 1、2、3 Up to the biggest 3 digit 999.

There's nothing to analyze

class Solution {
    public int[] printNumbers(int n) {
        int maxnum = (int) Math.pow(10, n)-1;
        int[] res = new int[maxnum];
        for (int i = 0; i < maxnum; i++) {
            res[i] = i+1;
        }
        return res;
    }
}

5. Delete the node of the linked list

Given the head pointer of one-way linked list and the value of a node to be deleted , Define a function to delete the node .

Return the head node of the deleted linked list .

analysis ：

Double pointer .

class Solution {
    public ListNode deleteNode(ListNode head, int val) {
        if (head.val == val){
            head = head.next;
            return head;
        } else {
            ListNode prenode = head;
            ListNode node = head.next;
            while(node!=null){
                if(node.val==val){
                    prenode.next = node.next;
                    break;
                }else {
                    prenode = node;
                    node = node.next;
                }
            }
        }
        return head;
    }
}