【重点】【DP】【耐心排序】300.最长递增子序列

题目
力扣题目链接
更好的方法是耐心排序，参见《算法小抄》的内容！！！

法1：DP

基础解法必须掌握！！！

Python

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        res, n = 1, len(nums)
        dp = [1] * n
        for i in range(n):
            for j in range(i):
                if nums[j] < nums[i]:
                    dp[i] = max(dp[i], dp[j] + 1)
                res = max(res, dp[i])
        return res

Java

class Solution {
    public int lengthOfLIS(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int maxLen = 1, n = nums.length;
        int[] dp = new int[n]; // 以i结尾的LIS
        Arrays.fill(dp, 1);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[i] > nums[j] && (dp[j] + 1 > dp[i])) {
                    dp[i] = dp[j] + 1;
                    maxLen = Math.max(maxLen, dp[i]);
                }
            }
        }

        return maxLen;
    }
}

法2：二分

参考：灵茶山艾府

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        g = []
        for x in nums:
            j = bisect_left(g, x)
            if j == len(g):
                g.append(x)
            else:
                g[j] = x
        return len(g)