LeetCode.532.K-diff Pairs in an Array 数组中的k-diff数对
题目描述
532 K-diff Pairs in an Array
https://leetcode-cn.com/problems/k-diff-pairs-in-an-array/
Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.
Example 1:
Input: [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:[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: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).
Note:
- The pairs (i, j) and (j, i) count as the same pair.
- The length of the array won’t exceed 10,000.
- All the integers in the given input belong to the range: [-1e7, 1e7].
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解题过程
用Map或者说哈希来实现 O(n)
分为k=0和k!=0两种情况
public int findPairs(int[] nums, int k) {
if (k < 0) {
return 0;
}
int count = 0;
// value -> 是否使用过
Map<Integer, Boolean> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
// 分为k=0和k!=0两种情况
if (0 != k) {
// 不重复处理元素
if (!map.containsKey(nums[i])) {
map.put(nums[i], false);
if (map.containsKey(nums[i] + k)) {
count++;
}
if (map.containsKey(nums[i] - k)) {
count++;
}
}
} else {
if (!map.containsKey(nums[i])) {
map.put(nums[i], false);
} else {
if (false == map.get(nums[i])) {
count ++;
map.put(nums[i], true);
}
}
}
}
return count;
}
GitHub代码
algorithms/leetcode/leetcode/_532_KDiffPairsInArray.java
https://github.com/masikkk/algorithms/blob/master/leetcode/leetcode/_532_KDiffPairsInArray.java
上一篇 LeetCode.617.Merge Two Binary Trees 合并二叉树
下一篇 LeetCode.530.Minimum Absolute Difference in BST 非负BST的最小绝对差
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