冏哥的程式之旅

 

public class LeetCode1365 {
	/*
	 * Easy
	 * 1365. How Many Numbers Are Smaller Than the Current Number
	 * Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it.
	 * That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].
	 * Return the answer in an array.
	 * 
	 * Example 1:
	 * Input: nums = [8,1,2,2,3]
	 * Output: [4,0,1,1,3]
	 * Explanation: 
	 * For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). 
	 * For nums[1]=1 does not exist any smaller number than it.
	 * For nums[2]=2 there exist one smaller number than it (1). 
	 * For nums[3]=2 there exist one smaller number than it (1). 
	 * For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).
	 * 
	 * Example 2:
	 * Input: nums = [6,5,4,8]
	 * Output: [2,1,0,3]
	 */
    public static void main(String[] args) {
        int[] input = {6,5,4,8};
        int[] res = smallerNumbersThanCurrent(input);
        for(int re : res) {
            System.out.print(re + ",");
        }
		
    }
	
    public static int[] smallerNumbersThanCurrent(int[] nums) {
    	int[] res = new int[nums.length];
    	for(int i = 0 ; i < nums.length ; i++) {
    		int count = 0;
    		for(int j = 0 ; j < nums.length ; j++) {
    			if(i == j) {
    				continue;
    			}
    			if(nums[i] > nums[j]) {
    				count++;
    			}
    		}
    		res[i] = count;
    	}
        return res;
    }
}