[LeetCode] Longest Increasing Subsequence

Given an unsorted array of integers, find the length of longest increasing subsequence.

For example,
Given [10, 9, 2, 5, 3, 7, 101, 18],
The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.

Your algorithm should run in O(n2) complexity.

Follow up: Could you improve it to O(n log n) time complexity?

解题思路

思路1：动态规划O( n2 )。i为递增子序列的末尾元素的位置，依次遍历i之前的位置，更新其值。
思路2：动态规划+二分搜索O(nlgn)。用一个附加数组保存递增序列的尾元素，依次遍历原数组中的元素，将其插入到附加数组中的正确位置，若插入最后一个元素之后，则更新最长递增子序列的长度。

实现代码

C++动态规划法实现代码如下:

// Runtime: 132ms
class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        if (nums.size() == 0)
        {
            return 0;
        }
        vector<int> cnt(nums.size(), 1);
        int res = 1;
        for (int i = 1; i < nums.size(); i++)
        {
            for (int j = 0; j < i; j++)
            {
                if (nums[j] < nums[i] && cnt[j] + 1 > cnt[i])
                {
                    cnt[i] = cnt[j] + 1;
                    res = max(res, cnt[i]);
                }
            }
        }

        return res;
    }
};

C++动态规划+二分法实现代码：

// Runtime: 4ms
class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        if (nums.size() == 0)
        {
            return 0;
        }
        vector<int> tail(nums.size(), 0);
        tail[0] = nums[0];
        int len = 1;
        for (int i = 1; i < nums.size(); i++)
        {
            int left = 0;
            int right = len - 1;
            while (left <= right)
            {
                int mid = (left + right) / 2;
                if (tail[mid] < nums[i])
                {
                    left = mid + 1;
                }
                else
                {
                    right = mid - 1;
                }
            }
            tail[left] = nums[i];
            if (left >= len)
            {
                len++;
            }
        }
        return len;
    }
};

java实现代码：

//Runtime: 2 ms
public class Solution {
    public int lengthOfLIS(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }
        int tail[] = new int[nums.length];
        tail[0] = nums[0];
        int len = 1;
        for (int i = 0; i < nums.length; i++) {
            int left = 0;
            int right = len - 1;
            while (left <= right) {
                int mid = (left + right) / 2;
                if (tail[mid] < nums[i]) {
                    left = mid + 1;
                }
                else {
                    right = mid - 1;
                }
            }
            tail[left] = nums[i];
            if (left == len) {
                ++len;
            }
        }

        return len;
    }
}