leetcode-Longest Increasing Subsequence

原创于 2015-11-04 12:03:25 发布 · 347 阅读

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本文介绍了一种求解最长递增子序列(LIS)问题的算法实现，通过动态规划方法给出了一个时间复杂度为O(n^2)的解决方案，并探讨了如何优化至O(n log n)的可能性。

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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?

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        if(nums.empty())
            return 0;
        int size=nums.size();
        int dp[size];
        fill(dp,dp+size,1);
        int maxLen=1;
        for(int i=1;i<size;++i){
            for(int j=0;j<i;++j){
                if(nums[i]>nums[j]&&dp[i]<dp[j]+1){
                    dp[i]=dp[j]+1;
                    maxLen=max(maxLen,dp[i]);
                }    
            }
        }
        return maxLen;
    }
};