300.leetcode-Longest Incresing Subsequence最长递增子序列

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?

方法一：采用动态规划的思想，定义一个一维数组dp[i],表示以nums[i]结尾的最长递增子序列。遍历nums[i]之前的数，如果发现某个数小于nums[i],则更新dp[i], dp[i]=max(dp[i], dp[j]+1).返回dp数组的最大值。时间复杂度为O(n2).

class Solution {public:    int lengthOfLIS(vector<int>& nums) {        int n = nums.size();if (n == 0)return 0;vector<int>dp(n, 1);int maxval=1;for (int i = 1; i<n; ++i){for (int j = i - 1; j >= 0; --j){if (nums[j] < nums[i])dp[i] = max(dp[i], dp[j] + 1);}if(dp[i]>maxval)maxval=dp[i];}return maxval;    }};

方法二：时间复杂度为O(nlogn).利用二分查找的思想。建立一个一维数组dp[i],开始遍历原数组，对每一个元素，用二分查找法找到第一个不小于它的的数字，如果这个数字不存在，则直接在后面添加该元素，如果存在，则将这个数字替换为当前遍历到的元素，最后返回dp的元素个数。注意：dp数组可能并不是一个真实的LIS.

int lengthOfLIS(vector<int>& nums) {    vector<int> res;    for(int i=0; i<nums.size(); i++) {        auto it = std::lower_bound(res.begin(), res.end(), nums[i]);        if(it==res.end()) res.push_back(nums[i]);        else *it = nums[i];    }    return res.size();}