leetcode 697
Given a non-empty array of non-negative integers nums
, the degree of this array is defined as the maximum frequency of any one of its elements.
Your task is to find the smallest possible length of a (contiguous) subarray of nums
, that has the same degree as nums
.
Example 1:
Input: [1, 2, 2, 3, 1]
Output: 2
Explanation:
The input array has a degree of 2 because both elements 1 and 2 appear twice.
Of the subarrays that have the same degree:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
The shortest length is 2. So return 2.
Example 2:
Input: [1,2,2,3,1,4,2]
Output: 6
给定非空整数nums的非空数组,该数组的度数被定义为其任何一个元素的最大频率。
你的任务是找到num的(连续的)子阵列的最小可能长度,其与nums具有相同的度数。
例1:
输入:[1,2,2,3,1]
输出:2
说明:
输入数组的度数为2,因为元素1和2都出现两次。
在具有相同程度的子阵列中:
[1,2,2,3,1],[1,2,2,3],[2,2,3,1],[1,2,2],[2,2,3],[2] ,2]
最短的长度是2.所以返回2。
例2:
输入:[1,2,2,3,1,4,2]
输出:6
class Solution {
public int findShortestSubArray(int[] nums) {
int maxcount = 1;
HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
for (int i : nums) {
if (hm.containsKey(i)) {
hm.put(i, hm.get(i) + 1);
if (maxcount < hm.get(i)) {
maxcount = hm.get(i);
}
} else {
hm.put(i, 1);
}
}
Set<Integer> set = hm.keySet();
int minlength = Integer.MAX_VALUE;
for (int s : set) {
int temp = Integer.MAX_VALUE;
if (hm.get(s) == maxcount) {
int i = 0, j = nums.length - 1;
while (nums[i] != s && i < j)
i++;
while (nums[j] != s && i < j)
j--;
temp = j - i + 1;
}
minlength = Math.min(temp, minlength);
}
return minlength;
}
}
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