Related Topics: “数组”: https://leetcode.com/tag/array/ “二分查找”: https://leetcode.com/tag/binary-search/ “动态规划”: https://leetcode.com/tag/dynamic-programming/ Similar Questions: “递增的三元子序列”: https://leetcode.com/problems/increasing-triplet-subsequence/ “俄罗斯套娃信封问题”: https://leetcode.com/problems/russian-doll-envelopes/ “最长数对链”: https://leetcode.com/problems/maximum-length-of-pair-chain/ “最长递增子序列的个数”: https://leetcode.com/problems/number-of-longest-increasing-subsequence/ “两个字符串的最小ASCII删除和”: https://leetcode.com/problems/minimum-ascii-delete-sum-for-two-strings/
Problem: Link to heading
给你一个整数数组 nums ，找到其中最长严格递增子序列的长度。
子序列是由数组派生而来的序列，删除（或不删除）数组中的元素而不改变其余元素的顺序。例如，[3,6,2,7] 是数组 [0,3,1,6,2,2,7] 的子序列。
示例 1：
输入：nums = [10,9,2,5,3,7,101,18]
输出：4
解释：最长递增子序列是 [2,3,7,101]，因此长度为 4 。
示例 2：
输入：nums = [0,1,0,3,2,3]
输出：4
示例 3：
输入：nums = [7,7,7,7,7,7,7]
输出：1
提示：
1 <= nums.length <= 2500-104 <= nums[i] <= 104
进阶：
- 你可以设计时间复杂度为
O(n2)的解决方案吗？ - 你能将算法的时间复杂度降低到
O(n log(n))吗?
Solution: Link to heading
//动态规划
int lengthOfLIS(vector<int>& nums) {
int n = nums.size();
if (n == 0) return 0;
vector<int> dp(n,0);
for (int i = 0; i < n; ++i){
dp[i] = 1;
for (int j = 0; j < i; ++j){
if(nums[j] < nums[i])
dp[i] = max(dp[j]+1, dp[i]);
}
}
return *max_element(dp.begin(), dp.end());
}
//贪心+二分查找
int lengthOfLIS(vector<int>& nums) {
int len = 1, n = nums.size();
if (n == 0) return 0;
vector<int> d(n+1,0);
d[len] = nums[0];
for (int i = 1; i < n; ++i) {
if(nums[i] > d[len]) {
d[++len] = nums[i];
}else{
int l = 1, r = len, pos = 0;
while(l <= r) {
int mid = (l + r) >> 1;
if(d[mid] < nums[i]) {
pos = mid;
l = mid + 1;
}else{
r = mid - 1;
}
}
d[pos + 1] = nums[i];
}
}
return len;
}
};