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;
	}
}