LeetCode 300 Longest Increasing Subsequence

題目描述

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.

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分析

定義一個數組s，s[i]用來定義從0…i的LIS，求解s[i+1]時，循環i次。時間複雜度是O(n2)。有更好的算法，能夠到O(nlogn)，過段時間研究+_+

O(n2)的僞代碼如下：

// 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

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代碼

    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，動態規劃求解
        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;
    }