Longest Increasing Subsequence

来源:转载

问题:

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 is4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.


思路:

    该题采用动态规划的思想来解决,用F(i)表示以数组i处元素结尾的最长递增子序列,则F(i + 1) = max(F(k) + 1), (其中k在0~i之间,且数组的第k个元素小于第i + 1 个元素)。


代码:

class Solution {public: int lengthOfLIS(vector<int>& nums) { int size = nums.size(); if(size < 1){ return 0; } int *cache = new int[size]; for(int i = 0; i < size; i++){ cache[i] = 0; } cache[0] = 1; int result = 1; for(int i = 1; i < size; i++){ int maxsize = 1; for(int j = i - 1; j >= 0; j--){ if(nums[i] > nums[j]){ if(cache[j] + 1 > maxsize){ maxsize = cache[j] + 1; } } } cache[i] = maxsize; if(cache[i] > result){ result = cache[i]; } } return result; }};





分享给朋友:
您可能感兴趣的文章:
随机阅读: