LeetCode.1207- only element occurrences (Unique Number of Occurrences)

This is the first Ogawa 419 update, the first 452 Pian original

Look problems and prepare

Introduced today is LeetCode algorithm in title Easy level of 269 questions (overall title number is 1207 ). Given an array of integers arr, if and only if the unique number of occurrences of each element in the array, return true.

E.g:

Input: arr = [1,2,2,1,1,3]
Output: true
Description: The value 1 appears 3 times, 2 times the value 2 occurs, the value of 1 appears 3 times. No two values are the same number of occurrences.

Input: arr = [1,2]
Output: false

Input: arr = [-3,0,1, -3,1,1,1, -3,10,0]
Output: true

Restrictions :

• 1 <= arr.length <= 1000
• -1000 <= arr[i] <= 1000
  
  

The first solution

The title means that the number of occurrences of elements in the array to determine whether or not the only, we can directly use HashMapand HashSetcombination of the first traversing arrelement, the element value as key, the number of occurrences as a valuedeposit HashMap, and then traverse HashMapall value, deposit HashSet, if The current valuename exists, return false.

public boolean uniqueOccurrences(int[] arr) {
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
    for (int num : arr) {
        map.put(num, map.getOrDefault(num, 0)+1);
    }
    Set<Integer> set = new HashSet<Integer>();
    for (Integer count : map.values()) {
        if (set.contains(count)) {
            return false;
        }
        set.add(count);
    }
    return true;
}

The second solution

Use intArray count. Because the subject and defines the length of the array element value range of the original, so we can resolve this problem by means of counting.

First count, a length of 2001 declare intarrays count, plus the value of the original array 1000 as a new index array. A second count, the same length as a declaration of the 2001 intarray count2, if countthe current element is not equal to 0, as will the current element count2array index value accumulation element, determines whether the addition was complete the element value is greater than or equal to 2, if it exceeds, direct return false.

public boolean uniqueOccurrences2(int[] arr) {
    int[] count = new int[2001];
    for (int num : arr) {
        count[num+1000]++;
    }
    int[] count2 = new int[2001];
    for (int con : count) {
        if (con != 0) {
            count2[con]++;
            if (count2[con] >= 2) {
                return false;
            }
        }
    }
    return true;
}

A third solution

And a second solution similar to the above, except that the second step count intarray into a booleantype, other thinking the same.

public boolean uniqueOccurrences3(int[] arr) {
    int[] count = new int[2001];
    for (int num : arr) {
        count[num+1000]++;
    }
    boolean[] flag = new boolean[2001];
    for (int con : count) {
        if (con > 0) {
            if (flag[con]) {
                return false;
            } else {
                flag[con] = true;
            }
        }
    }
    return true;
}

summary

Thematic algorithm has now updated LeetCode algorithm of feature articles 275 + articles, public Number dialog box reply [ data structures and algorithms ], [ algorithm ], [ data structures Any words] in obtaining a series of articles collection.

That's all, if you have any good solution ideas, suggestions or other issues, you can exchange comments below, thumbs up, leave a message, it is the greatest reward in watching and supporting me!