leetcode 673. Number of Longest Increasing Subsequence
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写在前面
contest 49最后一题。一道简单的动态规划题目,可是在做的时候总是想用贪婪算法解决,事实上是不可解的,另外,在求得最长连续序列后,我的想法是回溯求所有可能,最后导致超时,其实只需要在原先方法基础上修改即可,仍然是dp问题。
题目描述
Given an unsorted array of integers, find the number of longest increasing subsequence.
Example 1:
Input: [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequence are [1, 3, 4, 7] and [1, 3, 5, 7].
Example 2:
Input: [2,2,2,2,2]
Output: 5
Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences’ length is 1, so output 5.
Note: Length of the given array will be not exceed 2000 and the answer is guaranteed to be fit in 32-bit signed int.
思路分析
之所以第一时间想到的是贪婪算法,可能还是对题目的理解不够全面,认为第一个开始递增的数一定是序列开始(稍微一个极端的例子就可以说明这个思路是错的,比如序列 [1,2,-100,-99,….0] 按照贪婪的思路得到的长度是2,而事实上最长子序列是从-100到0),当然如果接触过LIS,会知道它的标准解法就是经典的动态规划,令dp[i]表示以nums[i]结尾的最长子序列,最后对dp的结果求max即可,我们先写出求最长递增序列的代码。
1.最长递增子序列
class Solution {public: int LengthOfLIS(vector<int>& nums) { // 就是一个dp vector<int> dp(nums.size(),1); int maxLen = 1; for(int i = 1;i<nums.size();++i) { for(int j = 0;j<i;++j) { if(nums[i]>nums[j]) { if(dp[i]<dp[j]+1){ dp[i] = dp[j]+1; } } } maxLen = max(maxLen,dp[i]); } return maxLen; }};
2.最长递增子序列的个数
我们知道上述解法,核心就是不断更新dp[i],从而得到当前的最优解。然而,最优解可能不止一个,这种情况会发生在 dp[i] == dp[j]+1,即i与j具有相同的序列长度,那么我们更新当前长度ct[i] = ct[i]+ct[j]若dp[j]+1>dp[i],则更新最长子序列的个数为dp[j]的个数,最终写出的代码如下:
class Solution {public: int findNumberOfLIS(vector<int>& nums) { // 就是一个dp vector<int> dp(nums.size(),1); vector<int> ct(nums.size(),1); int maxLen = 1; for(int i = 1;i<nums.size();++i) { for(int j = 0;j<i;++j) { if(nums[i]>nums[j]) { if(dp[i]<dp[j]+1){ dp[i] = dp[j]+1; ct[i] = ct[j]; } else if(dp[i] == dp[j]+1){ ct[i]+=ct[j]; } } } maxLen = max(maxLen,dp[i]); } int ret = 0; for(int i = 0;i<ct.size();++i) if(dp[i]==maxLen) ret+=ct[i]; return ret; }};
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