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
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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;
    }
};

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
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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
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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;
    }
};






posted @ 2016-12-09 23:22  copperface  阅读(206)  评论(0)  编辑  收藏  举报