300. Longest Increasing Subsequence“编程题”

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： O(n^2) dynamic programming
state equation, 不仅取决于前一时刻的状态，而是取决于前一段时间里的最小值或者最大值。常见的O(n^2)dp.

int lengthOfLIS(vector& nums) {
    int len = nums.size();
    int ans = 0;
    vector dp(len, 0);
    for (int i = 0; i < len; i++) {
        int maxval = 0;
        for (int j = 0; j < i; j++) {
            if (nums[i] > nums[j]) {
                maxval = max(maxval, dp[j]);
            }
        }
        dp[i] = maxval + 1;
        ans = max(ans, dp[i]);
    }
    return ans;
}