添加时间复杂度和空间复杂度

Author: soulmachine

C++/LeetCodet题解(C++版).pdf

@@ -41,6 +41,7 @@ \subsubsection{分析}
 \subsubsection{代码}
 \begin{Code}
 //LeetCode, Word Ladder
+// 时间复杂度O(n)，空间复杂度O(n)
 class Solution {
 public:
     typedef string state_t;
@@ -141,6 +142,7 @@ \subsubsection{代码}
 
 \begin{Code}
 //LeetCode, Word Ladder II
+// 时间复杂度O(n)，空间复杂度O(n)
 class Solution {
 public:
     vector<vector<string> > findLadders(string start, string end,
@@ -250,6 +252,7 @@ \subsubsection{代码}
 \begin{Code}
 // LeetCode, Surrounded Regions
 // BFS
+// 时间复杂度O(n)，空间复杂度O(n)
 class Solution {
 public:
     void solve(vector<vector<char>> &board) {
@@ -278,8 +281,8 @@ \subsubsection{代码}
         visit(board, i, j, q);
         while (!q.empty()) {
             int cur = q.front(); q.pop();
-            const int x = cur / board[0].size();
-            const int y = cur % board[0].size();
+            int x = cur / board[0].size();
+            int y = cur % board[0].size();
             visit(board, x - 1, y, q);
             visit(board, x, y - 1, q);
             visit(board, x + 1, y, q);

C++/chapBFS.tex

@@ -17,6 +17,7 @@ \subsubsection{代码}
 \begin{Code}
 //LeetCode, Pow(x, n)
 // 二分法，$x^n = x^{n/2} * x^{n/2} * x^{n\%2}$
+// 时间复杂度O(logn)，空间复杂度O(1)
 class Solution {
 public:
     double pow(double x, int n) {
@@ -58,6 +59,7 @@ \subsubsection{代码}
 \begin{Code}
 // LeetCode, Longest Substring Without Repeating Characters
 // 二分查找
+// 时间复杂度O(logn)，空间复杂度O(1)
 class Solution {
 public:
     int sqrt(int x) {

C++/chapDivideAndConquer.tex

@@ -22,10 +22,10 @@ \subsubsection{描述}
 OJ's undirected graph serialization:
 Nodes are labeled uniquely.
 
-We use \code{#} as a separator for each node, and \code{,} as a separator for node label and each neighbour of the node.
-As an example, consider the serialized graph \code{\{0,1,2#1,2#2,2\}}.
+We use \code{\#} as a separator for each node, and \code{,} as a separator for node label and each neighbour of the node.
+As an example, consider the serialized graph \code{\{0,1,2\#1,2\#2,2\}}.
 
-The graph has a total of three nodes, and therefore contains three parts as separated by \code{#}.
+The graph has a total of three nodes, and therefore contains three parts as separated by \code{\#}.
 \begin{enumerate}
 \item First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
 \item Second node is labeled as 1. Connect node 1 to node 2.
@@ -50,7 +50,7 @@ \subsubsection{分析}
 \subsubsection{代码}
 \begin{Code}
 // LeetCode, Clone Graph
-// DFS
+// DFS，时间复杂度O(n)，空间复杂度O(n)
 class Solution {
 public:
     UndirectedGraphNode *cloneGraph(const UndirectedGraphNode *node) {
@@ -80,7 +80,7 @@ \subsubsection{代码}
 
 \begin{Code}
 // LeetCode, Clone Graph
-// BFS
+// BFS，时间复杂度O(n)，空间复杂度O(n)
 class Solution {
 public:
     UndirectedGraphNode *cloneGraph(const UndirectedGraphNode *node) {

C++/chapGraph.tex

@@ -34,25 +34,21 @@ \subsubsection{分析}
 \subsubsection{代码}
 \begin{Code}
 // LeetCode, Jump Game
-// 思路一
+// 思路1，时间复杂度O(n)，空间复杂度O(1)
 class Solution {
 public:
     bool canJump(int A[], int n) {
-        int right_most = 0; // 最右能跳到哪里
-        for (int start = 0; start <= right_most; start++) {
-            if (A[start] + start > right_most)
-                right_most = A[start] + start;
-
-            if (right_most >= n - 1) return true;
-        }
-        return false;
+        int reach = 1; // 最右能跳到哪里
+        for (int i = 0; i < reach && reach < n; ++i)
+            reach = max(reach,  i + 1 + A[i]);
+        return reach >= n;
     }
 };
 \end{Code}
 
 \begin{Code}
 // LeetCode, Jump Game
-// 思路二
+// 思路2，时间复杂度O(n)，空间复杂度O(1)
 class Solution {
 public:
     bool canJump (int A[], int n) {
@@ -71,7 +67,7 @@ \subsubsection{代码}
 
 \begin{Code}
 // LeetCode, Jump Game
-// 思路三，动规
+// 思路三，动规，时间复杂度O(n)，空间复杂度O(n)
 class Solution {
 public:
     bool canJump(int A[], int n) {
@@ -117,6 +113,7 @@ \subsubsection{分析}
 \subsubsection{代码}
 \begin{Code}
 // LeetCode, Jump Game II
+// 时间复杂度O(n)，空间复杂度O(1)
 class Solution {
 public:
     int jump(int A[], int n) {
@@ -141,6 +138,30 @@ \subsubsection{代码}
 };
 \end{Code}
 
+\begin{Code}
+// LeetCode, Jump Game II
+// 时间复杂度O(n)，空间复杂度O(1)
+class Solution {
+public:
+    int jump(int A[], int n) {
+        int result = 0;
+        // the maximum distance that has been reached
+        int last = 0;
+        // the maximum distance that can be reached by using "ret+1" steps
+        int cur = 0;
+        for (int i = 0; i < n; ++i) {
+            if (i > last) {
+                last = cur;
+                ++result;
+            }
+            cur = max(cur, i + A[i]);
+        }
+
+        return result;
+    }
+};
+\end{Code}
+
 
 \subsubsection{相关题目}
 \begindot
@@ -166,6 +187,7 @@ \subsubsection{分析}
 \subsubsection{代码}
 \begin{Code}
 // LeetCode, Best Time to Buy and Sell Stock
+// 时间复杂度O(n)，空间复杂度O(1)
 class Solution {
 public:
     int maxProfit(vector<int> &prices) {
@@ -208,6 +230,7 @@ \subsubsection{分析}
 \subsubsection{代码}
 \begin{Code}
 // LeetCode, Best Time to Buy and Sell Stock II
+// 时间复杂度O(n)，空间复杂度O(1)
 class Solution {
 public:
     int maxProfit(vector<int> &prices) {
@@ -251,7 +274,7 @@ \subsubsection{分析}
 \subsubsection{代码}
 \begin{Code}
 // LeetCode, Longest Substring Without Repeating Characters
-// 贪心法
+// 时间复杂度O(n)，空间复杂度O(1)
 class Solution {
 public:
     int lengthOfLongestSubstring(string s) {
@@ -297,6 +320,7 @@ \subsubsection{分析}
 \subsubsection{代码}
 \begin{Code}
 // LeetCode, Container With Most Water
+// 时间复杂度O(n)，空间复杂度O(1)
 class Solution {
 public:
     int maxArea(vector<int> &height) {

C++/chapGreedy.tex

@@ -48,6 +48,7 @@
 \include{chapDivideAndConquer}
 \include{chapGreedy}
 \include{chapDynamicProgramming}
+\include{chapGraph}
 \include{chapImplement}
 
 \appendix % 开始附录，章用字母编号