Problem:

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?

即从一个无序数组中找出最长的递增子序列的长度.

Solution-1:

常规的DP.时间复杂度O(n^2)

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public class Solution {
    public int lengthOfLIS(int[] nums) {
        if(nums == null || nums.length == 0) return 0;
        int [] len = new int[nums.length];
        len[0] = 1;
        int max = 1;
        for(int i = 1;i<nums.length;++i){
            len[i] = 1;
            for(int j = i-1;j>=0;--j){
                if(nums[i] > nums[j]) len[i] = Math.max(len[i], len[j] +1);
            }
            max = Math.max(max, len[i]);
        }
        return max;
    }
}

Solution-2:

O(nlogn).很精妙(推荐),但很难理解.参考1, 参考2

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public class Solution {
    public int lengthOfLIS(int[] nums) {
        if(nums == null) return 0;
        if(nums.length <= 1) return nums.length;
        int [] tails = new int[nums.length];
        int size = 0;
        for(int i = 0;i<nums.length;++i){
            int l = 0, r = size;//在[0,size]之间搜索
            while(l<r){//其实就是在 tails 数组中找 nums[i]的 lower_bound
                int mid = l + (r-l)/2;
                if(tails[mid] > nums[i]) r = mid;
                if(tails[mid] < nums[i]) l = mid+1;
                else r = mid;
            }
            if(l==size) ++size;//如果tails中所有的元素都小于nums[i], 此时l==size
            tails[l] = nums[i];
        }
        return size;
    }
}

Reference: