今天完成了三道题目,总结一下:
1: Length of last word(细节实现题)
此题有一些细节需要注意(比如 “a_ _” 最后一个单词是a, 而不是遇到空格就直接算成没有),别的基本就是模拟了。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 class Solution { 2 public: 3 int lengthOfLastWord(const char *s) { 4 string str = s; 5 if(str.empty()) return 0; 6 int index; 7 // 从后向前扫描到第一个不是空格的字符 8 for(index = str.size()-1;index>=0;index--) 9 { 10 if(str[index]!=' ') 11 break; 12 } 13 if(index<0) return 0; 14 int i; 15 //从此字符开始,往前遇到空格停止计算或者把字符串扫描完 16 for(i=index;i>=0;i--) 17 { 18 if(str[i] == ' ') 19 break; 20 } 21 return (index-i); 22 } 23 };
2. Trapping Rain Water
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1]
, return 6
.
这道题目是直方图类题目,贴出的解法是book上的,两轮扫描,第一轮记录左前缀最大值和右前缀最大值; 第二轮扫描,对每个位置,可以积的水是
max(min(max_left,max_right)-height,0) height 是当前位置的格子高度。 这道题目应该算是一道技巧题了。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 class Solution { 2 public: 3 int trap(int A[], int n) { 4 if(n<2) return 0; 5 int* left_max = new int[n]; 6 int* right_max = new int[n]; 7 8 // Two pass algorithm 9 left_max[0] = A[0]; 10 right_max[n-1] = A[n-1]; 11 for(int i=1;i<n;i++) 12 { 13 left_max[i] = max(left_max[i-1],A[i]); 14 right_max[n-1-i] = max(right_max[n-i],A[n-1-i]); 15 } 16 int water = 0; 17 for(int i=0;i<n;i++) 18 { 19 water += max(min(left_max[i],right_max[i])-A[i],0); 20 } 21 return water; 22 23 } 24 };
3.Valid Parentheses
Given a string containing just the characters '('
, ')'
, '{'
, '}'
, '['
and ']'
, determine if the input string is valid.
The brackets must close in the correct order, "()"
and "()[]{}"
are all valid but "(]"
and "([)]"
are not.
简单的括号匹配题,一般都是用栈来实现。实现细节要注意,错误都犯在控制变量的初始化上了(test 变量每次set成true之后,没有重新初始化)
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 class Solution { 2 public: 3 bool isValid(string s) { 4 stack<int> st; 5 bool test=false;int num; 6 for(int i=0;i<s.size();i++) 7 { 8 switch (s[i]) 9 { 10 case '(': num = 1;st.push(num);break; 11 case '[': num = 2;st.push(num);break; 12 case '{': num = 3;st.push(num);break; 13 case ')': num = 1;test = true;break; 14 case ']': num = 2;test =true;break; 15 case '}': num = 3;test = true; break; 16 } 17 if(test) 18 { 19 if(st.empty() || st.top()!=num) return false; 20 else st.pop(); 21 test =false; 22 } 23 } 24 if(st.empty()) return true; 25 else return false; 26 } 27 };