Longest Harmonious Subsequence (E)
题目
We define a harmonious array as an array where the difference between its maximum value and its minimum value is exactly 1.
Given an integer array nums, return the length of its longest harmonious subsequence among all its possible subsequences.
A subsequence of array is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements.
Example 1:
Input: nums = [1,3,2,2,5,2,3,7]
Output: 5
Explanation: The longest harmonious subsequence is [3,2,2,2,3].
Example 2:
Input: nums = [1,2,3,4]
Output: 2
Example 3:
Input: nums = [1,1,1,1]
Output: 0
Constraints:
1 <= nums.length <= 2 * 10^4-10^9 <= nums[i] <= 10^9
题意
在给定数组中找到一个子数组，使其中的最大值和最小值的差正好为1。
思路
根据题意，这样的子数组中只可能存在两种不同的数，有以下两种方法：
先将数组排序，再利用滑动窗口找到首尾差为1的子数组，复杂度为(O(N))。
或可以用HashMap记录每个数字的出现次数，在遍历时，当前数为num，判断num-1和num+1是否存在，存在的话则比较max(ans, num出现次数 + (num-1) or (num+1)出现次数)。
代码实现
Java
排序
class Solution {
public int findLHS(int[] nums) {
if (nums.length == 1) {
return 0;
}
int ans = 0;
int left = 0, right = 1;
Arrays.sort(nums);
while (left <= nums.length - 2) {
if (nums[right] - nums[left] == 1) {
ans = Math.max(ans, right - left + 1);
right++;
} else if (nums[right] - nums[left] > 1) {
left++;
} else {
right++;
}
if (left == right) right++;
if (right == nums.length) break;
}
return ans;
}
}
Hash
class Solution {
public int findLHS(int[] nums) {
int ans = 0;
Map<Integer, Integer> map = new HashMap<>();
for (int num : nums) {
int cnt = map.getOrDefault(num, 0) + 1;
if (map.containsKey(num - 1)) {
ans = Math.max(ans, map.get(num - 1) + cnt);
}
if (map.containsKey(num + 1)) {
ans = Math.max(ans, map.get(num + 1) + cnt);
}
map.put(num, cnt);
}
return ans;
}
}