leetcode——TwoSum、3Sum

TwoSum

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].

解析：寻找序列中唯一的和为目标数的两个元素，并返回下标。做法一是直接暴力搜索，但是时间效率低下。另一种可行的做法是先将元素排序，然后用两个指针指向头尾，通过改变指针去寻找目标元素。由于需要返回下标，因此可以先用pair存储元素与其初始下标。具体实现代码如下，时间复杂度为O(nlogn)：

class Solution {public:static bool cmp(pair<int, int>& a, pair<int, int>& b){return a.first < b.first;}    vector<int> twoSum(vector<int>& nums, int target) {        vector<int> v;        vector<pair<int, int> >num;        int n = nums.size(), j = 0, k = n - 1;        for(int i=0; i<n; i++)        num.push_back(make_pair(nums[i], i));        sort(num.begin(), num.end(), cmp);        while(j < k)        {        if(num[j].first + num[k].first < target)j ++;        else if(num[j].first + num[k].first > target)k --;        else        {        v.push_back(num[j].second);        v.push_back(num[k].second);        break;        }}return v;    }};

还有一种实现比较简易，利用map容器存储元素，之后只需要对每个元素nums[i]判断是否存元素target-nums[i]即可，代码如下：

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        vector<int> ans;        map<int, int> M;        int n = nums.size();        for(int i=0; i<n; i++)        M[nums[i]] = i+1;        for(int i=0; i<n; i++)        if(M[target - nums[i]] != i+1 && M[target - nums[i]] > 0)        {        ans.push_back(i);        ans.push_back(M[target - nums[i]]-1);        break;}return ans;    }};

3Sum

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],A solution set is:[  [-1, 0, 1],  [-1, -1, 2]]

解析：根据TwoSum可以很容易得到3Sum的解法，排序之后，对每个元素nums[i]，在序列nums[i+1]到nums[n-1]中寻找和为0-nums[i]的两个元素。具体代码如下，时间复杂度为O(n^2)：

class Solution {public:    vector<vector<int> > threeSum(vector<int>& nums) {        vector<vector<int> > ans;        int n = nums.size(), m = -1;        sort(nums.begin(), nums.end());        for(int i=0; i<n; i++)        if(i == 0 || nums[i] != nums[i-1])        {        int j = i + 1, k = n - 1;        while(j < k)        {        if(nums[i] + nums[j] + nums[k] < 0)j ++;        else if(nums[i] + nums[j] + nums[k] > 0)k --;        else        {while(j+1 < k && nums[j+1]==nums[j])j ++;while(j < k-1 && nums[k-1]==nums[k])k --;        vector<int> v;        v.push_back(nums[i]);        v.push_back(nums[j]);        v.push_back(nums[k]);        ans.push_back(v);        j ++;        k --;        }}}return ans;    }};