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(DP LIS) leetcode 300. Longest Increasing Subsequence

Given an unsorted array of integers, find the length of longest increasing subsequence.

Example:

Input: [10,9,2,5,3,7,101,18]
Output: 4 
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. 

Note:

  • 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?

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参考官方题解:https://leetcode.com/articles/longest-increasing-subsequence/

这个题是最长上升子序列问题。先在某一个位置,姑且用i来表示,然后,判断nums[i]是否比前面的i数大?如果大,那么再比较dp[i] 和 dp[j] + 1 哪个比较大(j就是前面的数字的下标)。

C++代码:

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        if(nums.size() == 0) return 0;
        int n = nums.size();
        vector<int> dp(n,1);  //边界值就是1,因为无论如何LIS中的最小值就是1,否则就是0.
        int res = 1;
        for(int i = 1; i < n; i++){
            for(int j = 0; j < i; j++){
                if(nums[i] > nums[j] && dp[i] < dp[j] + 1)
                    dp[i] = dp[j] + 1;
            }
            res = max(res,dp[i]);
        }
        return res;
    }
};

 

posted @ 2019-05-10 21:01  PJCK  阅读(32)  评论(0编辑  收藏