leetcode上很多题目是一系列的,循序渐进,比如TwoSum,3Sum,4Sum的问题。
- TwoSum
Given an array of integers, find two numbers such that they add up to a specific target number. - 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. - 4Sum
Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
要求也很简单,找到给定数组的k个数,使得这k个数和为给定的数值target.
附加的则是一些计算index,求所有满足的组合,去重的操作等。
2Sum
先从最简单的TwoSum开始。
如果是杂乱无序的一组数,求和不得不要遍历所有的数值。因此第一想到的就是排序。
观察下排序后的数组
A < B < C < D
如果A + B = target,那么A+C,B+D都不可能等于target,也就是说只有B + C可能等于target。
因此可以这么计算:
while (left < right):
sum = a[left] + a[right]
if sum < target: #和太小,增大left.
++left
else if sum > target: #和太大,减小right.
--right
else #找到一组
++left
--right
写成代码如下:
vector<int> twoSum(vector<int> &numbers, int target)
{
vector<int> vec = numbers;
sort(numbers.begin(), numbers.end());
vector<int> ans;
int left = 0, right = numbers.size() - 1;
while (true)
{
int sum = numbers[left] + numbers[right];
if (sum > target)
--right;
else if (sum < target)
++left;
else
break;
}
//原题要求返回index,此处计算index.
int p = -1,q = -1;
for (int i = 0; i < vec.size(); ++i)
{
if (vec[i] == numbers[left] || vec[i] == numbers[right])
{
if (p < 0)
p = i;
else
q = i;
}
}
ans.push_back(p + 1);
ans.push_back(q + 1);
return ans;
}
3Sum
解决了TwoSum,3Sum就好理解了,数组排好序后,取定一个值,然后计算该值后面满足sum的两个数即可。
可以转化为解决TwoSum的问题。
注意多了一个去重的要求,因为是排好序的数组,去重只要做到临近的数字相同的话略过即可。
代码如下:
vector<vector<int> > threeSum(vector<int> &num)
{
vector<vector<int> > ans;
if (num.size() < 3)
return ans;
sort(num.begin(), num.end());
for (unsigned int i = 0; i < num.size(); ++i)
{
if (i != 0 && num[i] == num[i-1])
continue;
int target = -num[i];
int left = i + 1, right = num.size() - 1;
while (left < right)
{
int sum = num[left] + num[right];
if (sum < target)
{
++left;
}
else if (sum > target)
{
--right;
}
else
{
vector<int> v;
v.push_back(num[i]);
v.push_back(num[left]);
v.push_back(num[right]);
ans.push_back(v);
++left;
while (left < right && num[left] == num[left - 1])
++left;
--right;
while (left < right && num[right] == num[right + 1])
--right;
}
}
}
return ans;
}
4Sum以及kSum
到这里其实发现kSum不过是一个递归的问题,要求k个数之和为target[k],可以转化为k-1个数之和target[k-1]。
代码如下:
//make sure num is sorted in non-descending order before this function.
vector<vector<int> > kSum(vector<int> &num, int left, int target, int k)
{
vector<vector<int> > ans;
if (num.size() < k)
return ans;
if (k == 2)
{
int i = left, j = num.size() - 1;
while (i < j)
{
int sum = num[i] + num[j];
if (sum < target)
{
++i;
}
else if (sum > target)
{
--j;
}
else
{
vector<int> v;
v.push_back(num[i]);
v.push_back(num[j]);
ans.push_back(v);
++i;
while (i < j && num[i] == num[i-1])
++i;
--j;
while (i < j && num[j] == num[j + 1])
--j;
}
}
return ans;
}
for (unsigned int i = left; i <= num.size() - k; ++i)
{
if (i != left && num[i] == num[i - 1])
continue;
vector<vector<int> > vec = kSum(num, i + 1, target - num[i], k - 1);
for (unsigned int j = 0; j < vec.size(); ++j)
{
vector<int> v = vec[j];
v.insert(v.begin(), num[i]);
ans.push_back(v);
}
}
return ans;
}
vector<vector<int> > fourSum(vector<int> &num, int target)
{
sort(num.begin(), num.end());
return kSum(num, 0, target, 4);
}
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