java 532.阵列中的K-diff对（第1个）.java

Posted 2021-05-08

import java.util.Map.Entry;

public class Solution {
    public int findPairs(int[] nums, int k) {
		if (nums == null || nums.length == 0 || k<0)	return 0;
		
		Arrays.sort(nums);
		int cnt = 0, left = 0, right = 1;
	
		while (left < nums.length && right < nums.length)	{
			//每次檢查指標是否重疊，以及距離小於K移動右邊指標
			if (left == right || nums[left] + k > nums[right])
				right++;
			//反之距離大於K移動左邊指標
			else if (nums[left] + k < nums[right])
				left++;
			else	{
				//找到符合的區間
				cnt++;
				left++;	//移動指標
				//檢查是否有重複直存在
				while (left < nums.length && nums[left] == nums[left-1])	left++;
				right++;
			}
				
		}
		
		return cnt;
	
    
    }
}

public class Solution {
    public int findPairs(int[] nums, int k) {
        if (nums == null || nums.length < 1 || k < 0) return 0;
        Map<Integer, Integer> map = new HashMap<>();
        for (int num : nums) {
            map.put(num, map.getOrDefault(num, 0) + 1);
        }
        int res = 0;
        if (k == 0) {
            for (int value : map.values()) {
                if (value > 1) res++;
            }
            return res;
        } else if(k > 0){
            for (int num : map.keySet()) {
                if(map.containsKey(num + k)) res++;
            }
            return res;
        }
        return res;
    }
}

public class Solution {
    public int findPairs(int[] nums, int k) {
        if(k < 0) return 0;
        int lower = Integer.MIN_VALUE;
        Arrays.sort(nums);
        int i = 0, j = 0;
        int diff = 0;
        int count = 0;
        while( i < nums.length && j < nums.length) {
            diff = nums[j] - nums[i];
            if(diff == k && i != j) {
                if(nums[i] > lower) {
                    count++;
                    lower = nums[i];
                }
                i++;
            } else if(diff > k) i++;
            else j++;
        }
        return count;
    }
}