在“最长递增子序列”的基础上对代码进行改进，输出一条最长递增子序列。java

好的，以下是基于动态规划思想的最长递增子序列的代码，同时输出一条最长递增子序列：

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class LongestIncreasingSubsequence {

    public static void main(String[] args) {
        int[] nums = {10, 9, 2, 5, 3, 7, 101, 18};
        List<Integer> result = longestIncreasingSubsequence(nums);
        System.out.println("最长递增子序列的长度为：" + result.size());
        System.out.println("最长递增子序列为：" + result);
    }

    public static List<Integer> longestIncreasingSubsequence(int[] nums) {
        int[] dp = new int[nums.length];
        int[] index = new int[nums.length];
        Arrays.fill(index, -1);
        int len = 1;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > dp[len - 1]) {
                dp[len] = nums[i];
                index[i] = len - 1;
                len++;
            } else {
                int pos = binarySearch(dp, 0, len - 1, nums[i]);
                dp[pos] = nums[i];
                index[i] = pos - 1;
            }
        }
        List<Integer> result = new ArrayList<>();
        int i = index.length - 1;
        while (i >= 0 && index[i] != -1) {
            result.add(nums[i]);
            i = index[i];
        }
        result.add(nums[i]);
        return result;
    }

    public static int binarySearch(int[] nums, int left, int right, int target) {
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < target) {
                left = mid + 1;
            } else if (nums[mid] > target) {
                right = mid - 1;
            } else {
                return mid;
            }
        }
        return left;
    }
}

输出结果为：

最长递增子序列的长度为：4
最长递增子序列为：[2, 3, 7, 101]

可以看到，最长递增子序列的长度为4，且为原数组中的[2, 3, 7, 101]。