300. Longest Increasing Subsequence

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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?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

首先是复杂度O(n^2)的动态规划算法。建立一个数组存放前i位的最长升序子序列，遍历0-j，如果nums[j]小于nums[i]，用math.max(dp[j] + 1, dp[i])代替dp[i]。代码如下：

class Solution {public:    // There's a typical DP solution with O(N^2) Time and O(N) space     // DP[i] means the result ends at i    // So for dp[i], dp[i] is max(dp[j]+1), for all j < i and nums[j] < nums[i]    int lengthOfLIS(vector<int>& nums) {        const int size = nums.size();        if (size == 0) { return 0; }         vector<int> dp(size, 1);        int res = 1;        for (int i = 1; i < size; ++i) {            for (int j = 0; j < i; ++j) {                if (nums[j] < nums[i]) {                    dp[i] = max(dp[i], dp[j]+1);                }            }            res = max (res, dp[i]);        }        return res;    }};

第二种是复杂度O(nlogn)的二分查找方法。建立数组存放可能存在的升序子序列。从头遍历nums，试图把每一个num放到dp相应的位置里，最后返回dp存下的最长升序自序列的长度。代码如下：

public class Solution {    public int lengthOfLIS(int[] nums) {        int[] dp = new int[nums.length];        int size = 0;        for (int x: nums) {            int i = -1, j = size;            while (i + 1 < j) {                int mid = i + (j - i) / 2;                if (dp[mid] < x) {                    i = mid;                } else {                    j = mid;                }            }            if (i < 0) {                dp[0] = x;            } else {                dp[i + 1] = x;            }            if (i + 1 == size) {                dp[size] = x;                size ++;            }        }        return size;    }}

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