LeetCode 300. Longest Increasing Subsequence 解题报告

300. Longest Increasing Subsequence

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Question

Total Accepted: 17302  Total Submissions: 51952  Difficulty: Medium

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.

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    十分典型的DP问题！

    定义一个f函数，f(i)表示以第一个元素为结尾的最长递增子序列的长度。要求f(i),那么得先求出f(0),f(1)......f(n-1)。

    我的AC代码如下：

public class LongestIncreasingSubsequence {

	public static void main(String[] args) {
		int[] a = { 10, 9, 2, 5, 3, 7, 101, 18 };
		System.out.println(lengthOfLIS(a));
	}

	public static int lengthOfLIS(int[] nums) {
		if(nums.length == 0) return 0;
		int[] f = new int[nums.length];
		for (int i = 0; i < f.length; i++) {
			f[i] = 1;
		}
		int max = 1;
		for (int i = 1; i < nums.length; i++) {
			for (int j = 0; j < i; j++) {
				if (nums[i] > nums[j] && f[j] + 1 > f[i]) {
					f[i] = f[j] + 1;
					if (f[i] > max)
						max = f[i];
				}
			}
		}

		return max;
	}
}