C语言刷题 LeetCode 30天挑战 （七）哈希计数法

//Counting Elements

//Given an integer array arr ,count element x such that x + 1 is also in arr

//lf there're duplicates in arr , count them seperately.

//Example 1:

//Input: arr =1,2,3

//0utput:2

//Explanation:1 and 2 are counted cause 2 and 3 are in arr.

//Example 2:

//Input: arr =1,1,3,3,5,5,7,7

//0utput:0

//Explanation: No numbers are counted, cause there's no 2,4,6, or 8 in arr.

//Example 3:

//Input: arr = 1,3,2,3,5,8

//Output:3

//Explanation:0,1 and 2 are counted cause 1,2 and 3 are in arr.

//Example 4:

//Input: arr =1,1,2,2

//0utput:2

//Explanation:Two is are counted cause 2 is in arr.

#include <stdio.h>

int countElements(int *arr, int arrSize) {
    // Boolean array to mark presence of elements in the input array.
    int elementExists[1002] = {0}; // Supports values from 0 to 1000

    // Mark the presence of elements in the array
    for (int i = 0; i < arrSize; ++i) {
        elementExists[arr[i]] = 1;
    }

    int returnNum = 0;
    // Check if `arr[i] + 1` exists in the array
    for (int i = 0; i < arrSize; ++i) {
        if (elementExists[arr[i] + 1]) {
            returnNum++;
        }
    }

    return returnNum;
}

int main()
{
    int arr[] = {1,1,2,2};
    printf("%d",countElements(arr,sizeof(arr)/sizeof(arr[0])));
    return 0;
}