leetcode 300. Longest Increasing Subsequence-优快云博客

本文介绍了一种求解最长递增子序列(LIS)问题的动态规划算法，该算法能在O(n²)的时间复杂度内解决问题，并探讨了如何进一步优化至O(nlogn)的时间复杂度。通过具体示例，展示了状态转移方程的应用。

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300. 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?

动态规划

状态 state[i]表示前i个数字中以第i个结尾的LIS的长度
初始化：以每一位结尾都初始化为1
状态转移方程：state[i] = max(state[j] + 1)，其中 j < i && num[j] <= num[i]
结果：max(state[i], i从0到n-1)

public class Solution {
    public int lengthOfLIS(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }

        int[] state = new int[nums.length];
        int max = 0;
        for (int i = 0; i < nums.length; i++) {
            state[i] = 1;
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    state[i] = state[i] > state[j] + 1 ? state[i] : state[j] + 1;
                }
            }
            if (state[i] > max) {
                max = state[i];
            }
        }

        return max;
    }
}