300. Longest Increasing Subsequence

0300-longest-increasing-subsequence/memo.md

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+### step1
+
+O(N ^ 2)の動的計画法しか思いつかなかった。
+
+### step2
+
+ChatGPTに聞いて長さ i+1 の増加部分列の「末尾の最小値」をvectorでもっておけばいいことがわかった。
+これでO(nlogn)になる。
+
+### step3
+
+3回通すまで書き直し。

0300-longest-increasing-subsequence/step1.cpp

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+class Solution {
+public:
+    int lengthOfLIS(vector<int>& nums) {
+        map<int, int> val_and_length;
+
+        int max_length = 1;
+        for (int i = 0; i < nums.size(); i++) {
+            val_and_length[nums[i]] = max(1, val_and_length[nums[i]]);
+            for (auto it = val_and_length.begin(); it != val_and_length.end(); it++) {
+                auto [val, length] = *it;
+                if (nums[i] > val) {
+                    val_and_length[nums[i]] = max(length + 1, val_and_length[nums[i]]);
+                    max_length = max(max_length, val_and_length[nums[i]]);
+                }
+            }
+        }
+        return max_length;
+    }
+};

0300-longest-increasing-subsequence/step2.cpp

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+class Solution {
+public:
+    int lengthOfLIS(vector<int>& nums) {
+        vector<int> length_and_val;
+
+        for (int x : nums) {
+            auto it = lower_bound(length_and_val.begin(), length_and_val.end(), x);
+
+            if (it == length_and_val.end()) {
+                length_and_val.push_back(x);
+            } else {
+                *it = x;
+            }
+        }
+
+        return length_and_val.size();
+    }
+};

0300-longest-increasing-subsequence/step3.cpp

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+class Solution {
+public:
+    int lengthOfLIS(vector<int>& nums) {
+        vector<int> length_and_val;
+
+        for (int x : nums) {
+            auto it = lower_bound(length_and_val.begin(), length_and_val.end(), x);
+
+            if (it == length_and_val.end()) {
+                length_and_val.push_back(x);
+            } else {
+                *it = x;
+            }
+        }
+
+        return length_and_val.size();
+    }
+};