[LeetCode] #1# Two Sum : 数组/哈希表/二分查找/双指针

一. 题目
1. Two Sum
Total Accepted: 241484 Total Submissions: 1005339 Difficulty: Easy

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

 
二. 题意
  • 给定一个数组和一个目标值
  • 找出数组中两个成员，两者之和为目标值
  • 假设一定存在一个解

 

三. 分析

  • 算法核心:
    • 三种方法:
      • 暴力搜索: O(n^2): 超时
      • 哈希表: O(n)
      • 先快排, 后二分查找: O(nlogn) + O(nlogn)
      • 先快排, 后使用双指针分别指向数组头和尾，同时双向遍历数组: O(nlogn) + O(n)
  • 实现细节:
    • 算法逻辑相对简单
    • 实现细节相对容易

 

四. 题解

  • 哈希表
    • 实现　　
     1 class Solution {
 2 public:
 3     vector<int> twoSum(vector<int>& nums, int target) {
 4         vector<int> res;
 5         unordered_map<int, int> m;
 6         
 7         for (int i = 0; i < nums.size(); i++) m[nums[i]] = i;
 8         
 9         for (int i = 0; i < nums.size(); i++) {
10             if (m.count(target - nums[i]) && m[target - nums[i]] != i) {
11                 res.push_back(i);
12                 res.push_back(m[target - nums[i]]);
13                 return res;
14             }
15         }
16         
17         return res;
18     }
19 };
    • 结果

　　

　　　　　　

  • 快排-二分查找
    • 实现
 1 class Solution {
 2 public:
 3     vector<int> twoSum(vector<int>& B, int target) {
 4         vector<int> res;
 5         vector<pair<int, int>> A;
 6         
 7         for (int i = 0; i < B.size(); i++) {
 8             A.push_back(make_pair(B[i], i));
 9         }
10         
11         my_qsort(A, 0, A.size() - 1);
12         
13         for (int i = 0; i <= A.size(); i++) {
14             int left = i + 1, right = A.size() - 1;
15             while (left <= right) {
16                 int mid = left + (right - left) / 2;
17                 if (A[mid].first == target - A[i].first) {
18                     res.push_back(A[i].second);
19                     res.push_back(A[mid].second);
20                     return res;
21                 }
22                 if (A[mid].first < target - A[i].first) left = mid + 1;
23                 else right = mid - 1;
24             }
25         }
26         
27         return res;
28     }
29 private:
30     void my_qsort(vector<pair<int, int>>& A, int l, int r) {
31         if (l > r) return;
32         
33         pair<int, int> key = A[l];
34         int nl= l, nr = r;
35         while (l < r) {
36             pair<int, int> tmp;
37             while (A[r].first >= key.first && l < r) r--;
38             while (A[l].first <= key.first && l < r) l++;
39             
40             
41             tmp = A[l];
42             A[l] = A[r];
43             A[r] = tmp;
44         }
45         A[nl] = A[l];
46         A[l] = key;
47         
48         my_qsort(A, nl, l - 1);
49         my_qsort(A, l + 1, nr);
50     }
51 };
    • 结果　　

　　　　　

 

  • 快排-双指针
    • 实现
 1 class Solution {
 2 public:
 3     vector<int> twoSum(vector<int>& B, int target) {
 4         vector<int> res;
 5         vector<pair<int, int>> A;
 6         
 7         for (int i = 0; i < B.size(); i++) {
 8             A.push_back(make_pair(B[i], i));
 9         }
10         
11         int left = 0, right = A.size() - 1;
12         my_qsort(A, 0, A.size() - 1);
13         
14         while (left < right) {
15             if (A[left].first + A[right].first < target) left++;
16             else if (A[left].first + A[right].first > target) right--;
17             else {res.push_back(A[left].second), res.push_back(A[right].second); return res;};
18         }
19         
20         return res;
21     }
22 private:
23     void my_qsort(vector<pair<int, int>>& A, int l, int r) {
24         if (l > r) return;
25         
26         pair<int, int> key = A[l];
27         int nl= l, nr = r;
28         while (l < r) {
29             pair<int, int> tmp;
30             while (A[r].first >= key.first && l < r) r--;
31             while (A[l].first <= key.first && l < r) l++;
32             
33             
34             tmp = A[l];
35             A[l] = A[r];
36             A[r] = tmp;
37         }
38         A[nl] = A[l];
39         A[l] = key;
40         
41         my_qsort(A, nl, l - 1);
42         my_qsort(A, l + 1, nr);
43     }
44 };
    • 结果

　　　　　

 

五. 相似题

posted on 2016-06-08 19:56  zhongyuansh  阅读(353)  评论(0)  编辑  收藏  举报

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