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& a, pair& b)
{
return a.first < b.first;
}
vector twoSum(vector& nums, int target) {
vector v;
vector > num;
int n = nums.size(), j = 0, k = n - 1;
for(int i=0; i 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 twoSum(vector& nums, int target) {
vector ans;
map M;
int n = nums.size();
for(int i=0; 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] ]
class Solution {
public:
vector > threeSum(vector& nums) {
vector > ans;
int n = nums.size(), m = -1;
sort(nums.begin(), nums.end());
for(int i=0; i 0) k --;
else
{
while(j+1 < k && nums[j+1]==nums[j]) j ++;
while(j < k-1 && nums[k-1]==nums[k]) k --;
vector v;
v.push_back(nums[i]);
v.push_back(nums[j]);
v.push_back(nums[k]);
ans.push_back(v);
j ++;
k --;
}
}
}
return ans;
}
};