LeetCode300. Longest Increasing Subsequence-优快云博客

本文介绍了解决LeetCode 300题——最长递增子序列的方法，通过动态规划实现，给出了一种O(n^2)的时间复杂度解决方案，并提供了完整的C++代码实现。

LeetCode300. Longest Increasing Subsequence

题目：

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.

Follow up: Could you improve it to O(n log n) time complexity?

题意分析：一般是这种类型的关于长度的题目，很容易就想到了动态分析去。于是本题也是往动态分析去看的。

vector<int> dp(nums.size(), 1);
int res = 0;
for (int i = 0; i < nums.size(); ++i) {
	for (int j = 0; j < i; ++j) {
		if (nums[i] > nums[j]) {
			dp[i] = max(dp[i], dp[j] + 1);
	}
	}
res = max(res, dp[i]);
}

以下是代码部分：

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        vector<int> dp(nums.size(), 1);
		int res = 0;
		for (int i = 0; i < nums.size(); ++i) {
			for (int j = 0; j < i; ++j) {
				if (nums[i] > nums[j]) {
					dp[i] = max(dp[i], dp[j] + 1);
				}
			}
			res = max(res, dp[i]);
		}
		return res;
    }
};