# [LintCode] Longest Increasing Subsequence 最长递增子序列

Given a sequence of integers, find the longest increasing subsequence (LIS).

You code should return the length of the LIS.

Have you met this question in a real interview?

Example

For [5, 4, 1, 2, 3], the LIS  is [1, 2, 3], return 3

For [4, 2, 4, 5, 3, 7], the LIS is [4, 4, 5, 7], return 4

Challenge

Time complexity O(n^2) or O(nlogn)

Clarification

What's the definition of longest increasing subsequence?

* The longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence's elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous, or unique.

* https://en.wikipedia.org/wiki/Longest_common_subsequence_problem

class Solution {
public:
/**
* @param nums: The integer array
* @return: The length of LIS (longest increasing subsequence)
*/
int longestIncreasingSubsequence(vector<int> nums) {
vector<vector<int> > solutions;
longestIncreasingSubsequence(nums, solutions, 0);
int res = 0;
for (auto &a : solutions) {
res = max(res, (int)a.size());
}
return res;
}
void longestIncreasingSubsequence(vector<int> &nums, vector<vector<int> > &solutions, int curIdx) {
if (curIdx >= nums.size() || curIdx < 0) return;
int cur = nums[curIdx];
vector<int> best_solution;
for (int i = 0; i < curIdx; ++i) {
if (nums[i] <= cur) {
best_solution = seqWithMaxLength(best_solution, solutions[i]);
}
}
vector<int> new_solution = best_solution;
new_solution.push_back(cur);
solutions.push_back(new_solution);
longestIncreasingSubsequence(nums, solutions, curIdx + 1);
}
vector<int> seqWithMaxLength(vector<int> &seq1, vector<int> &seq2) {
if (seq1.empty()) return seq2;
if (seq2.empty()) return seq1;
return seq1.size() < seq2.size() ? seq2 : seq1;
}
};

http://www.cnblogs.com/lishiblog/p/4190936.html

http://blog.xiaohuahua.org/2015/01/26/lintcode-longest-increasing-subsequence/

posted @ 2015-10-19 12:04  Grandyang  阅读(5566)  评论(0编辑  收藏  举报