subject

Ideas

1. Violent words arrays.sort Then you can find it directly , But such words seem to have O(nlogn) Time complexity of dissatisfaction O(n), Since it is O(n) The algorithm of , That's a traversal , It can't be dp,dp Well, that's right O(n^2)
2. The solution of the problem is really simple , But I don't know why, but I didn't expect , Even if I think of it, I don't think it counts O(n)… Take note .

Code

    public int longestConsecutive(int[] nums) {
    
        //hashset Duplicate data storage is not allowed   Direct filtration nums Data in 
        Set<Integer> num_set = new HashSet<Integer>();
        // Join in 
        for(int num:nums) num_set.add(num);
        // The answer returned 
        int ans=0;
        for (int num : num_set) {
    
            // For every number   Find the former and the latter 
            if (!num_set.contains(num - 1)) {
    
                // Be sure to find the smallest in the sequence   Therefore, it is possible to become the smallest without the previous one 
                int currentNum = num;
                // Maximum possible length 
                int currentStreak = 1;
                // Move back 
                while (num_set.contains(currentNum + 1)) {
    
                    currentNum += 1;
                    currentStreak += 1;
                }
                // Just compare 
                ans = Math.max(ans, currentStreak);
            }
        }
        return ans;
    }