题目描述

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.

分析

定义一个数组s，s[i]用来定义从0…i的LIS，求解s[i+1]时，循环i次。时间复杂度是O(n2)。有更好的算法，能够到O(nlogn)，过段时间研究+_+

O(n2)的伪代码如下：

// Let S[i] be OPT(i)
S[1] := 1
L := S[1]
for i from 2 to n
    S[i] := 1 // at least contain A[i]
    for j from 1 to i-1
        if A[j] < A[i] then S[i] := max( S[i], S[j]+1 )
    end
    L := max( L, S[i] )
end
return L

代码

    public int lengthOfLIS(int[] nums) {

        if (nums == null || nums.length == 0) {
            return 0;
        }

        int n = nums.length;

        int max = 1;

        int[] s = new int[n];
        Arrays.fill(s, 1);

        // 自底向上，动态规划求解
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[j] < nums[i]) {
                    s[i] = Math.max(s[i], s[j] + 1);
                }
            }
            max = Math.max(max, s[i]);
        }

        return max;
    }