540. Single element in ordered array / 1684 Count the number of consistent strings

540. A single element in an ordered array 【 Medium question 】【 A daily topic 】

Ideas ：

1. I compare dishes , When writing, I didn't expect how logn Complexity , Just look straight from front to back , My complexity is n, Problem solving binary search is very good , To study the .
2. Define the target number cur, The initial value defaults to the first element value , Define count variables cnt, The initial value is 0.
3. Traverse nums, If cnt Less than 2, Then it means that this number pair has not been counted , Then if the current number is equal to cur, that cnt++, If not equal to cur, So it means that cur The number has no matching number （ because nums Is ordered , The current element is no longer equal to cur 了 , Then the latter is more unlikely to equal cur）; If cnt Not less than 2, Then it must be at least equal to 2 Of , Then it will be explained at this time cur Its matching number has been found , So will cur Update to current element ,cnt Updated to 1.
4. In special cases, when nums The length is 1 when , Then it's impossible to for The answer is returned inside the loop , So you should return externally at this time nums The value of this unique element in .

Code ：

class Solution {
    
    public int singleNonDuplicate(int[] nums) {
    
        int cur = nums[0],cnt = 0;
        for (int num : nums) {
    
            if (cnt<2){
    
                if (num == cur){
    
                    cnt++;
                }else {
    
                    return cur;
                }
            }else {
    
                cur = num;
                cnt = 1;
            }
        }
        return cur;
    }
}

when ：

when 1ms, If the time complexity does not meet the requirements, it will not be posted .

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1684. Count the number of consistent strings 【 Simple questions 】

Ideas ：

1. First define a hash set take allowed The characters are stored , Then define int Type variable ans, The initial value is 0.
2. Traversal string array words Every string of word, Traverse the current word Every character of , If the current character is in set Does not exist in the , So similar flag bit flag（ The default is true） Set as false And exit the character cycle of the current string . If flag by true, Then the current string word And allowed Strings are similar , that ans++.
3. Finally back to ans that will do .

Code ：

class Solution {
    
    public int countConsistentStrings(String allowed, String[] words) {
    
        Set<Character> set = new HashSet<>();
        for (char ch : allowed.toCharArray()){
    
            set.add(ch);
        }
        int ans = 0;
        for (String word : words){
    
            boolean flag = true;
            for (char ch : word.toCharArray()) {
    
                if (!set.contains(ch)){
    
                    flag = false;
                    break;
                }
            }
            if (flag){
    
                ans++;
            }
        }
        return ans;
    }
}

when ：

At present, there is no official solution , Time is not fast either , Make do with it .