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?

O(n²) complexity 的解法：定义f[i], f[i] 表示到数组下标为i的最长递增子序列的长度。f[i] = max(f[i], f[j] + 1) 其中(0 <= j < i ,且nums[i] >= nums[j]).

 1 public class Solution {
 2     public int lengthOfLIS(int[] nums) {
 3         if (nums == null || nums.length == 0) {
 4             return 0;
 5         }
 6         int max = 0;
 7         int[] f = new int[nums.length];
 8         Arrays.fill(f, 1);
 9         for (int i = 0; i < nums.length; i++) {
10             for (int j = 0; j < i; j++) {
11                 if (nums[i] >= nums[j]) {
12                     f[i] = Math.max(f[i], f[j] + 1);
13                 }
14             }
15             max = Math.max(max, f[i]);
16         }
17         return max;
18     }
19 }

Longest Increasing Subsequence

原文：http://www.cnblogs.com/FLAGyuri/p/5339877.html