【Leetcode】217. Contains Duplicate
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方法一:
思路:
用一个set存储数组中出现过的元素,遍历数组元素,若该元素已存在于set中,则返回true,否则将其加入set。
public class Solution { public boolean containsDuplicate(int[] nums) { Set<Integer> set = new HashSet<Integer>(); int len = nums.length; for (int i = 0; i < len; i++) { if (set.contains(nums[i])) return true; else set.add(nums[i]); } return false; }}
Runtime:21ms
方法二:
思路:
先将数组排序,然后从第一个开始遍历,如果和后一个值相等,则返回true,终止。
public class Solution { public boolean containsDuplicate(int[] nums) { Arrays.sort(nums); int len = nums.length; for (int i = 0; i < len - 1; i++) { if (nums[i] == nums[i + 1]) return true; } return false; }}
时间复杂度为O(nlogn),空间复杂度为O(1)
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