Sum 系列题解
Two Sum题解
题目来源：https://leetcode.com/problems/two-sum/description/
Description
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, and you may not use the same element twice.
Example
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
Solution
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* twoSum(int* nums, int numsSize, int target) {
int* result = (int*)malloc(2 * sizeof(int));
int i, j;
for (i = 0; i < numsSize; i++) {
for (j = 0; j < numsSize; j++) {
if (i == j) continue;
if (nums[i] + nums[j] == target) {
if (i < j) {
result[0] = i;
result[1] = j;
} else {
result[0] = j;
result[1] = i;
}
return result;
}
}
}
return result;
}
解题描述
这道题目还是比较简单的，为了找到目标数字的下标，使用的是直接用双层循环遍历数组里面任意两个数的和，检查和是否等于给定的target。之后再返回存有所求的两个数字的下标的数组。
更优解法
2018.1.24 更新：
之前这道题的做法属于暴力破解，时间复杂度还是较高的，达到了O(n^2)，查了一些资料之后发现使用哈希可以把时间复杂度降到O(n)：
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> hash;
int size = nums.size();
vector<int> res(2);
for (int i = 0; i < size; i++) {
auto got = hash.find(target - nums[i]);
if (got != hash.end()) {
res[0] = got -> second;
res[1] = i;
return res;
}
hash[nums[i]] = i;
}
}
};
3Sum 题解
题目来源：https://leetcode.com/problems/3sum/description/
Description
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.
Example
For example, given array S = [-1, 0, 1, 2, -1, -4],
A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]
Solution
class Solution {
private:
vector<vector<int> > twoSum(vector<int>& nums, int end, int target) {
vector<vector<int> > res;
int low = 0;
int high = end;
while (low < high) {
if (nums[low] + nums[high] == target) {
vector<int> sum(2);
sum[0] = nums[low++];
sum[1] = nums[high--];
res.push_back(sum);
// 去重
while (low < high && nums[low] == nums[low - 1])
low++;
while (low < high && nums[high] == nums[high + 1])
high--;
} else if (nums[low] + nums[high] > target) {
high--;
} else {
low++;
}
}
return res;
}
public:
vector<vector<int> > threeSum(vector<int>& nums) {
vector<vector<int>> res;
int size = nums.size();
if (size < 3)
return res;
sort(nums.begin(), nums.end());
for (int i = size - 1; i >= 2; i--) {
if (i < size - 1 && nums[i] == nums[i + 1]) // 去重
continue;
auto sum2 = twoSum(nums, i - 1, 0 - nums[i]);
if (!sum2.empty()) {
for (auto& sum : sum2) {
sum.push_back(nums[i]);
res.push_back(sum);
}
}
}
return res;
}
};
解题描述
这道题是Two Sum的进阶，解法上采用的是先求Two Sum再根据求到的sum再求三个数和为0的第三个数，不过题意要求不一样，Two Sum要求返回数组下标，这道题要求返回具体的数组元素。而如果使用与Two Sum相同的哈希法去做会比较麻烦。
这里求符合要求的2数之和用的方法是，先将数组排序之后再进行夹逼的办法。并且为了去重，需要在2sum和3sum都进行去重。这样2sum夹逼时间复杂度为O(n)，总的时间复杂度为O(n^2)。
4Sum 题解
题目来源：https://leetcode.com/problems/4sum/description/
Description
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.
Note: The solution set must not contain duplicate quadruplets.
Example
For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.
A solution set is:
[
[-1, 0, 0, 1],
[-2, -1, 1, 2],
[-2, 0, 0, 2]
]
Solution
class Solution {
private:
vector<vector<int> > twoSum(vector<int>& nums, int end, int target) {
vector<vector<int> > res;
int low = 0;
int high = end;
while (low < high) {
if (nums[low] + nums[high] == target) {
vector<int> sum(2);
sum[0] = nums[low++];
sum[1] = nums[high--];
res.push_back(sum);
// 去重
while (low < high && nums[low] == nums[low - 1])
low++;
while (low < high && nums[high] == nums[high + 1])
high--;
} else if (nums[low] + nums[high] > target) {
high--;
} else {
low++;
}
}
return res;
}
vector<vector<int> > threeSum(vector<int>& nums, int end, int target) {
vector<vector<int>> res;
if (end < 2)
return res;
sort(nums.begin(), nums.end());
for (int i = end; i >= 2; i--) {
if (i < end && nums[i] == nums[i + 1]) // 去重
continue;
auto sum2 = twoSum(nums, i - 1, target - nums[i]);
if (!sum2.empty()) {
for (auto& sum : sum2) {
sum.push_back(nums[i]);
res.push_back(sum);
}
}
}
return res;
}
public:
vector<vector<int> > fourSum(vector<int>& nums, int target) {
vector<vector<int>> res;
int size = nums.size();
if (size < 4)
return res;
sort(nums.begin(), nums.end());
for (int i = size - 1; i >= 2; i--) {
if (i < size - 1 && nums[i] == nums[i + 1]) // 去重
continue;
auto sum3 = threeSum(nums, i - 1, target - nums[i]);
if (!sum3.empty()) {
for (auto& sum : sum3) {
sum.push_back(nums[i]);
res.push_back(sum);
}
}
}
return res;
}
};
解题描述
这道题可以说是3Sum的再次进阶，使用的方法和3Sum基本相同，只是在求3个数之和之后再套上一层循环。时间复杂度为O(n^3)。
3Sum Closest 题解
题目来源：https://leetcode.com/problems/3sum-closest/description/
Description
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
Example
For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
Solution
class Solution {
private:
inline int abInt(int val) {
return val >= 0 ? val : -val;
}
public:
int threeSumClosest(vector<int>& nums, int target) {
int size = nums.size();
if (size < 3)
return 0;
sort(nums.begin(), nums.end());
int closest = nums[0] + nums[1] + nums[2];
int first, second, third, sum;
for (first = 0; first < size - 2; ++first) {
if (first > 0 && nums[first] == nums[first - 1])
continue; // 去重
second = first + 1;
third = size - 1;
while (second < third) {
sum = nums[first] + nums[second] + nums[third];
if (sum == target)
return sum;
if (abInt(sum - target) < abInt(closest - target))
closest = sum;
if (sum < target)
++second;
else
--third;
}
}
return closest;
}
};
解题描述
这道题可以说是3Sum的变形，题意上是要求在数组里面找到三个数的和最接近target，也就是说可能不等于target。算法上面与3Sum的解法有一定的相似度，先固定一个位置first，在其后进行夹逼求最接近target的三数之和。