标签：right nums int list subsequence longest increasing left

300. Longest Increasing Subsequence Medium

Given an integer array nums, return the length of the longest strictly increasing subsequence.

A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].

Example 1:

Input: nums = [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.

Example 2:

Input: nums = [0,1,0,3,2,3]
Output: 4

Example 3:

Input: nums = [7,7,7,7,7,7,7]
Output: 1 

Constraints:

• 1 <= nums.length <= 2500
• -104 <= nums[i] <= 104

Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity?

 解法1:

class Solution {
    public int lengthOfLIS(int[] nums) {
        int[] dp = new int[nums.length];
        int result = 1;
        Arrays.fill(dp, 1);
        for(int i=0;i<nums.length-1;i++){
            for(int j=i+1;j<nums.length;j++){
                if(nums[i] < nums[j]) dp[j] = Math.max(dp[j], dp[i]+1);
                result = Math.max(result, dp[j]);
            }
        }
        return result;
    }
}

 时间复杂度： O(N²)

解法2:

class Solution {
    public int lengthOfLIS(int[] nums) {
        List<Integer> list = new ArrayList();
        for(int i=0;i<nums.length;i++){
            int pos = binarySearch(list, nums[i]);
            if(pos >= list.size()) list.add(nums[i]);
            else list.set(pos, nums[i]);
        }
        return list.size();
    }
    private int binarySearch(List<Integer> list, int target){
        int left = 0,right = list.size();
        while(left < right){
            int mid = left+(right-left)/2;
            if(list.get(mid) >= target) right = mid;
            else left = mid+1;
        }
        return left;
    }
}

 时间复杂度：O(NlogN)

 

标签：right,nums,int,list,subsequence,longest,increasing,left
来源： https://www.cnblogs.com/cynrjy/p/16598307.html

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