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  • 532. K-diff Pairs in an Array -- Medium

    Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.

    A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:

    • 0 <= i, j < nums.length
    • i != j
    • a <= b
    • b - a == k

    Example 1:

    Input: nums = [3,1,4,1,5], k = 2
    Output: 2
    Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
    Although we have two 1s in the input, we should only return the number of unique pairs.
    

    Example 2:

    Input: nums = [1,2,3,4,5], k = 1
    Output: 4
    Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
    

    Example 3:

    Input: nums = [1,3,1,5,4], k = 0
    Output: 1
    Explanation: There is one 0-diff pair in the array, (1, 1).
    

    Example 4:

    Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3
    Output: 2
    

    Example 5:

    Input: nums = [-1,-2,-3], k = 1
    Output: 2
    

    Constraints:

    • 1 <= nums.length <= 104
    • -107 <= nums[i] <= 107
    • 0 <= k <= 107

    two pointers, time = O(n), space = O(n)

    class Solution {
        public int findPairs(int[] nums, int k) {
            if (nums == null || nums.length == 0 || k < 0) {
                return 0;
            }
            Map<Integer, Integer> map = new HashMap<>();
            int count = 0;
            for (int i : nums) {
                map.put(i, map.getOrDefault(i, 0) + 1);
            }
            
            for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
                if (k == 0) {
                    if (entry.getValue() >= 2) {
                        count++;
                    } 
                } else {
                    if (map.containsKey(entry.getKey() + k)) {
                        count++;
                    }
                }
            }
            return count;
        }
    }
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/13770464.html
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