LeetCode 219. Contains Duplicate II
2017-09-01 15:22
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问题描述:
Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.分析问题:
给定一个数组,给定一个距离k。如果存在两个数相等,其下标的间隔小于等于k,就返回true。否者 返回false。代码实现:
public boolean containsNearbyDuplicate(int[] nums, int k) { if (nums == null || nums.length <= 1) { return false; } Map<Integer, Integer> keyLocationsMap = new HashMap<Integer, Integer>(); for (int i = 0; i < nums.length; i++) { if (keyLocationsMap.get(nums[i]) != null) { if ((i - keyLocationsMap.get(nums[i])) <= k) { return true; } } keyLocationsMap.put(nums[i], i); } return false; }
问题改进:
在上面解法中,使用了一个hashmap来存放一个数的值和下标,当值相等是,然后比较下标。如果下标超过了需要等下下标。可以使用一个hashSet,值存放当前下标之前的k个数,之前的数全部删除。代码实现:
public boolean containsNearbyDuplicate(int[] nums, int k) { Set<Integer> containsSet = new HashSet<Integer>(k ); for (int i = 0; i < nums.length; i++) { if (i > k) {//k+1的时候删除0. containsSet.remove(nums[i - k - 1]); } if (!containsSet.add(nums[i])) { return true; } } return false; }
总结
将一个问题限定在一定的范围内,设定边界是一种很好的方法。相关文章推荐
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