300. Longest Increasing Subsequence 解题记录

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.

解题思路：

这题可以采用动态规划的办法,新建一个等长数组count来记录子数组的最长长度,遍历数组nums,第二层依次遍历count数组得到当前元素的最长数组长度.

代码：

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int ret = 0, n = nums.size();
        vector<int> count(n, 1);
        for(int i = 0; i < n; i++){
            for(int j = 0; j < i; j ++){
                if(nums[i]>nums[j] && count[j]>=count[i])
                //当前元素大于遍历元素 且 遍历元素代表的最长长度+1大于当前元素长度
                    count[i] = count[j]+1;
            }
            ret = ret>count[i]?ret:count[i];  //每次得到当前元素的最长长度时与ret进行比较，毕竟当前元素的最长长度不代表所有元素中的最长长度
        }
        return ret;
    }
};