更新leetcode 300-319未上锁算法

Author: blacksea3

cpp/leetcode/300. longest-increasing-subsequence.cpp

@@ -1,5 +1,6 @@
 #include "public.h"
 
+/*
 //44ms, 71.18%
 //DP problem
 //O(n^2), 理论上可以结合二分查找降至O(nlongn), 待优化
@@ -30,3 +31,25 @@ class Solution {
 		}
 	}
 };
+*/
+
+//4ms, 96.72%
+//DP+二分
+//维持一个有序变长DP序列: 内部已排序, 长度表示当前最长上升子序列(但内容却不一定完全合法)
+//对每个新的num, 如果当前插入DP位置在末尾则DP长度+1,并插入, 否则替换掉可以插入的位置
+//最终返回DP的长度
+
+class Solution {
+public:
+	int lengthOfLIS(vector<int>& nums) {
+		vector<int> dp;
+
+		for (auto& num : nums)
+		{
+			vector<int>::iterator iter = lower_bound(dp.begin(), dp.end(), num);
+			if (iter == dp.end()) dp.emplace_back(num);
+			else dp[iter - dp.begin()] = num;
+		}
+		return dp.size();
+	}
+};

cpp/leetcode/301. remove-invalid-parentheses.cpp

@@ -1,13 +1,13 @@
 #include "public.h"
 
-//20ms, 46.26%
+//0ms, 100%
 //先统计最少需要去掉的左括号和右括号, 然后递归回溯
 
 class Solution {
 private:
 	unordered_set<string> res;
 	void recurse(const string& s, int index, int left_count, int right_count,
-		int left_rem, int right_rem, string expr)
+		int left_rem, int right_rem, string& expr)
 	{
 		if (index == s.size())
 		{
@@ -17,11 +17,12 @@ class Solution {
 		else
 		{
 			//直接丢弃当前的括号, 向内递归
-			if ((s[index] == '(' && left_rem > 0) || (s[index] == ')' && right_rem > 0))
-			{
-				recurse(s, index + 1, left_count, right_count, left_rem - (s[index] == '('),
-					right_rem - (s[index] == ')'), expr);
-			}
+			if ((s[index] == '(' && left_rem > 0))
+				recurse(s, index + 1, left_count, right_count, left_rem - 1,
+					right_rem, expr);
+			if ((s[index] == ')' && right_rem > 0))
+				recurse(s, index + 1, left_count, right_count, left_rem,
+					right_rem - 1, expr);
 
 			//记录当前内容
 			expr.push_back(s[index]);
@@ -56,7 +57,8 @@ class Solution {
 				if (left == 0) right++;
 				else left--;
 
-		recurse(s, 0, 0, 0, left, right, "");
+		string exp = "";
+		recurse(s, 0, 0, 0, left, right, exp);
 		vector<string> returnres(res.begin(), res.end());
 		return returnres;
 	}
@@ -66,7 +68,7 @@ class Solution {
 int main()
 {
 	Solution* s = new Solution();
-	vector<string> res = s->removeInvalidParentheses("(a)())()");
+	vector<string> res = s->removeInvalidParentheses("()())()");
 	return 0;
 }
 */

cpp/leetcode/303. range-sum-query-immutable.cpp

@@ -1,36 +1,20 @@
 #include "public.h"
 
-//36ms, 98.20%
+//36ms, 94.90%
 //设置一个vector, 应当存入下标0至当前下标的和
 
 class NumArray {
 private:
 	vector<int> vnums;
 public:
 	NumArray(vector<int>& nums) {
-		vnums.reserve(nums.size());
-		int sum = 0;
-		for (int i = 0; i < nums.size(); ++i)
-		{
-			sum += nums[i];
-			vnums[i] = sum;
-		}
+		vnums.reserve(nums.size() + 1);
+		vnums[0] = 0;
+		for (int i = 1; i <= nums.size(); ++i)
+			vnums[i] = vnums[i - 1] + nums[i - 1];
 	}
 
 	int sumRange(int i, int j) {
-		if (i == 0)
-		{
-			return vnums[j];
-		}
-		else
-		{
-			return vnums[j] - vnums[i - 1];
-		}
+		return vnums[j + 1] - vnums[i];
 	}
 };
-
-/**
- * Your NumArray object will be instantiated and called as such:
- * NumArray* obj = new NumArray(nums);
- * int param_1 = obj->sumRange(i,j);
- */

cpp/leetcode/304. range-sum-query-2d-immutable.cpp

@@ -1,6 +1,6 @@
 #include "public.h"
 
-//32ms, 91.48%
+//24ms, 98.46%
 //和上一题差不多一个意思
 //变成4个矩形
 
@@ -14,13 +14,6 @@ class NumMatrix {
 		vector<int> temp(matrix[0].size() + 1);
 		prev.resize(matrix.size() + 1, vector<int>(matrix[0].size() + 1, 0));
 
-		/*
-		for (int i = 0; i < matrix[0].size(); ++i)
-			prev[0][i] = 0;
-		for (int i = 1; i < matrix.size(); ++i)
-			prev[i][0] = 0;
-		*/
-
 		for (int row = 1; row <= matrix.size(); ++row)
 		{
 			int presum = 0;
@@ -36,9 +29,3 @@ class NumMatrix {
 		return prev[row2 + 1][col2 + 1] - prev[row2 + 1][col1] - prev[row1][col2 + 1] + prev[row1][col1];
 	}
 };
-
-/**
- * Your NumMatrix object will be instantiated and called as such:
- * NumMatrix* obj = new NumMatrix(matrix);
- * int param_1 = obj->sumRegion(row1,col1,row2,col2);
- */

cpp/leetcode/306. additive-number.cpp

@@ -1,14 +1,15 @@
 #include "public.h"
 
-//4ms, 91.58%
+//4ms, 75.96%
 //DFS
 
 class Solution {
 public:
 	bool isAdditiveNumber(string num) {
-		if (num.size() <= 2) return false;
-		for (int i = 1; 2 * i <= num.size(); i++) {
-			for (int j = 1; i + 2 * j <= num.size(); j++) {
+		int nSize = num.size();
+		if (nSize <= 2) return false;
+		for (int i = 1; 2 * i <= nSize; i++) {
+			for (int j = 1; i + 2 * j <= nSize; j++) {
 				if (dfs(num, str2int(num.substr(0, i)), str2int(num.substr(i, j)), i + j, j))
 					return true;
 			}
@@ -17,10 +18,11 @@ class Solution {
 		return false;
 	}
 private:
-	bool dfs(const string& num, long pre, long prepre, int i, int len) {
-		if (i == num.size()) return true;
-		if ((i > num.size())) return false;
-		for (int j = len; j <= num.size(); j++) {
+	bool dfs(string const& num, long pre, long prepre, int i, int len) {
+		int nSize = num.size();
+		if (i == nSize) return true;
+		if (i > nSize) return false;
+		for (int j = len; j <= nSize; j++) {
 			long cur = str2int(num.substr(i, j));
 			if (cur == -1) return false;
 			if (cur == pre + prepre && dfs(num, prepre, cur, i + j, j))
@@ -29,10 +31,10 @@ class Solution {
 		return false;
 	}
 
-	long str2int(const string& s) {
+	long str2int(string const& s) {
 		if (s.size() > 1 && s[0] == '0') return -1;
 		long sum = 0;
-		for (auto c : s) {
+		for (auto& c : s) {
 			if (sum >= LONG_MAX / 10) return -1;
 			sum = sum * 10 + c - '0';
 		}

cpp/leetcode/307. range-sum-query-mutable.cpp

@@ -1,6 +1,75 @@
 #include "public.h"
 
-//224ms, 24.73%, 待优化
+//40ms, 91.86%
+//分段求和, 伪线段树
+
+class NumArray {
+private:
+	vector<int> sums;
+	vector<int> cpNums;
+	int segSize;
+	int sSize;
+
+public:
+	NumArray(vector<int>& nums) {
+		int nSize = nums.size();
+		if (nSize == 0) return;
+
+		cpNums = nums;
+		segSize = (int)sqrt(nSize);
+		sSize = nSize / segSize;
+		sums.resize(sSize);
+		//初始化sums: 下标i 存放 nums中的[segSize*(i-1): segSize*i)的和
+		for (int index = 0; index < sSize; ++index)
+		{
+			sums[index] = accumulate(nums.begin() + segSize * index,
+				nums.begin() + segSize * (index + 1),
+				0);
+		}
+	}
+
+	void update(int i, int val) {
+		//求i的segSize的整数倍数, 向下取整
+		int mulIndex = i / segSize;
+		if (mulIndex != sSize) sums[mulIndex] += val - cpNums[i];
+		cpNums[i] = val;
+	}
+
+	int sumRange(int i, int j) {
+		//按区间求和
+		//求i/j的segSize整数倍数, 向上/下取整
+		//sti是实际连续求和段的起始sums坐标, enj是结束坐标
+		int muli, mulj, sti, enj;
+		int res = 0;
+
+		muli = i / segSize;
+		if (i%segSize != 0)  //例如: segSize:3, i:1, j=7; 一共是 012 345 678
+		{
+			res += accumulate(cpNums.begin() + i, cpNums.begin() + (muli + 1)*segSize, 0);
+			sti = muli + 1;
+		}
+		else sti = muli;
+
+		mulj = j / segSize;
+		if ((j + 1) % segSize != 0)
+		{
+			res += accumulate(cpNums.begin() + mulj * segSize, cpNums.begin() + j + 1, 0);
+			enj = mulj - 1;
+		}
+		else enj = mulj;
+
+		if (sti == enj + 2) return res - sums[enj + 1];
+		else if (sti <= (enj + 1))
+		{
+			return res + accumulate(sums.begin() + sti, sums.begin() + enj + 1, 0);
+		}
+		else return -99999; //dump
+	}
+};
+
+/*
+//224ms, 24.73%
+//纯暴力
 //vector<int>保存下标0至当前的元素和
 
 class NumArray {
@@ -32,10 +101,17 @@ class NumArray {
 		else return sums[j] - sums[i - 1];
 	}
 };
+*/
 
-/**
- * Your NumArray object will be instantiated and called as such:
- * NumArray* obj = new NumArray(nums);
- * obj->update(i,val);
- * int param_2 = obj->sumRange(i,j);
- */
+/*
+int main()
+{
+	vector<int> nums = { 1,3,5 };
+	NumArray* n = new NumArray(nums);
+	cout << n->sumRange(0, 2) << endl;  //9
+	n->update(1, 2);
+	cout << n->sumRange(0, 2) << endl;  //8
+	cout << n->sumRange(0, 1) << endl;  //3
+	cout << n->sumRange(0, 0) << endl;  //1
+}
+*/

cpp/leetcode/309. best-time-to-buy-and-sell-stock-with-cooldown.cpp

@@ -1,6 +1,6 @@
 #include "public.h"
 
-//4ms, 93.19%
+//4ms, 87.41%
 //DP problem
 //三个状态: 最大收益: 买入/可买/冷却
 //更新: 下一个买入=max(买入, 可买-价格), 下一个冷却=买入+价格, 下一个可买=max(可买, 冷却)

cpp/leetcode/310. minimum-height-trees.cpp

@@ -1,6 +1,6 @@
 #include "BinaryTree.h"
 
-//92ms, 97.39%
+//92ms, 84.24%
 //初始化将叶节点(nodes[i].size()==1)记录下来,然后对于叶节点的下一个节点,判断其删除那个叶节点后是否是叶节点,是则记录下来
 //直至当前只有1个或两个节点,他们就是root
 
@@ -32,11 +32,11 @@ class Solution {
 				int temp = q.front();
 				q.pop();
 				n--;
-				unordered_set<int>::iterator iter = nodes[temp].begin();
-				nodes[*iter].erase(temp);
-				if (nodes[*iter].size() == 1)
+				int beginIndex = *nodes[temp].begin();
+				nodes[beginIndex].erase(temp);
+				if (nodes[beginIndex].size() == 1)
 				{
-					next.push(*iter);
+					next.push(beginIndex);
 				}
 			}
 			q = next;
@@ -45,7 +45,7 @@ class Solution {
 		vector<int> res;
 		while (!q.empty())
 		{
-			res.push_back(q.front());
+			res.emplace_back(q.front());
 			q.pop();
 		}
 		return res;

cpp/leetcode/312. burst-balloons.cpp

@@ -1,6 +1,6 @@
 #include "public.h"
 
-//20ms, 98.42%
+//16ms, 98.95%
 //DP problem
 //将nums两端扩展: 例如3 1 5 8 扩展为 1 3 1 5 8 1
 //定义dp[i][j]表示戳破下标:[i,j]后的最大收益

cpp/leetcode/313. super-ugly-number.cpp

@@ -1,6 +1,6 @@
 #include "public.h"
 
-//76ms, 76.53%
+//64ms, 85.26%
 //DP problem
 //记录k个dp值, 表示下一个
 
@@ -9,12 +9,8 @@ class Solution {
 	int nthSuperUglyNumber(int n, vector<int>& primes) {
 		int k = primes.size();
 		vector<int> res(n, 0);
-		//int *res = new int[n];
 		res[0] = 1;
-		vector<int> pos(k, 0);
-		//int **pos = new int *[k];
-		for (int i = 0; i < k; i++)  //初始化k个下标
-			pos[i] = 0;
+		vector<int> pos(k, 0); //初始化k个下标成0
 		int next = 1;  //下一个存放位置
 		while (next < n)
 		{