18. 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.

Note: The solution set must not contain duplicate quadruplets.

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]
]

分析

两层循环，然后按照3Sum的做。O(n3)的复杂度。150ms

class Solution {

public:

vector<vector<int>> fourSum(vector<int>& nums, int target) {

int len = nums.size();

vector<vector<int> > result;

if(len < 4) return result;

sort(nums.begin(), nums.end());

for(int i = 0; i < len - 3; ++i){

if( i > 0 && nums[i] == nums[i - 1]){

continue;

}

for(int j = i + 1; j < len - 2; ++j){

if(j > i + 1 && nums[j] == nums[j - 1]){

continue;

}

int l = j + 1;

int r = len - 1;

while( l < r){

if( nums[i] + nums[j] + nums[l] + nums[r] == target){

result.push_back(vector<int>{nums[i], nums[j], nums[l], nums[r]});

l++;

while( l < r && nums[l] == nums[l - 1])

l++;

}

else if( nums[i] + nums[j] + nums[l] + nums[r] < target){

l++;

}

else{

r--;

}

return result;

}

};

进一步优化：https://discuss.leetcode.com/topic/37878/c-implementation-with-carefully-pruning-accelerates-a-lot-from-100ms-to-16ms

对排序后的数组，进行计算的时候，如果满足

1. 最靠前的4个数字之和 > target，则退出计算，因为以后也一定不会满足

2. 前面的数组和末尾的数字之和 < target，那么说明肯定中间数字之和一定是 < target的，继续

1 2	`if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target)` `break;` `if(nums[i]+nums[len-3]+nums[len-2]+nums[len-1]<target)` `continue;`

同时优化满足条件的判断顺序以及后续处理，可以得到如下代码：16ms

class Solution {

public:

vector<vector<int>> fourSum(vector<int>& nums, int target) {

int len = nums.size();

vector<vector<int> > result;

if(len < 4) return result;

sort(nums.begin(), nums.end());

for(int i = 0; i < len - 3; ++i){

if( i > 0 && nums[i] == nums[i - 1]){

continue;

}

/** cut edge to accelerate the speed **/

if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break;

if(nums[i]+nums[len-3]+nums[len-2]+nums[len-1]<target) continue;

for(int j = i + 1; j < len - 2; ++j){

if(j > i + 1 && nums[j] == nums[j - 1]){

continue;

}

/** cut edge to accelerate the speed **/

if(nums[i]+nums[j]+nums[j+1]+nums[j+2]>target) break;

if(nums[i]+nums[j]+nums[len-2]+nums[len-1]<target) continue;

int l = j + 1;

int r = len - 1;

while( l < r){

int sum = nums[i] + nums[j] + nums[l] + nums[r];

if( sum < target){

l++;

}

else if( sum > target){

r--;

}

else{

result.push_back(vector<int>{nums[i], nums[j], nums[l], nums[r]});

/** cut edge to accelerate the speed **/

r--;l++;

while( l < r && nums[l] == nums[l - 1])l++;

while( l < r && nums[r] == nums[r + 1])r--;

}

return result;

}

};

来自为知笔记(Wiz)

时间： 2024-08-06 07:55:53

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