How to Find the Closest Sum of Three in an Array using Two Point

  • 时间:2020-09-25 11:32:47
  • 分类:网络文摘
  • 阅读:95 次

Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example:
Given array nums = [-1, 2, 1, -4], and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

The straightforward solution has to be the bruteforce algorithm, that exhausts every three numbers using O(N^3) loop – which is obviously too slow.

Two Pointer Algorithm in O(nlogN)

We first need to sort the entire array which takes O(nlogN). Once we have determined the first two numbers in O(N^2), we can search the rest in (logN) as the array is sorted. The following algorithm takes O(n*n*logN).

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(begin(nums), end(nums));
        int sum = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < nums.size() - 2; ++ i) {
            for (int j = i + 1; j < nums.size() - 1; ++ j) {
                int k = nums.size() - 1;
                while (j < k) {
                    int cur = nums[i] + nums[j] + nums[k];
                    if (cur == target) {
                        return target;
                    } else if (cur < target) {
                        j ++;
                    } else {
                        k --;
                    }
                    if (abs(cur - target) < abs(sum - target)) {
                        sum = cur;
                    }
                }
            }
        }
        return sum;
    }
};
class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(begin(nums), end(nums));
        int sum = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < nums.size() - 2; ++ i) {
            for (int j = i + 1; j < nums.size() - 1; ++ j) {
                int k = nums.size() - 1;
                while (j < k) {
                    int cur = nums[i] + nums[j] + nums[k];
                    if (cur == target) {
                        return target;
                    } else if (cur < target) {
                        j ++;
                    } else {
                        k --;
                    }
                    if (abs(cur - target) < abs(sum - target)) {
                        sum = cur;
                    }
                }
            }
        }
        return sum;
    }
};

However, we don’t need to bruteforce the second number. Once the first number is settled, we can using two pointer algorithm to determine the remainding two numbers. If at anytime, we find a sum that is equal to the target, we immediately return the sum otherwise, we need to iteratively store the minimal sum difference.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(begin(nums), end(nums));
        int sum = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < nums.size() - 2; ++ i) {
            int j = i + 1;
            int k = nums.size() - 1;
            while (j < k) {
                int cur = nums[i] + nums[j] + nums[k];
                if (cur == target) {
                    return target;
                } else if (cur < target) {
                    j ++;
                } else {
                    k --;
                }
                if (abs(cur - target) < abs(sum - target)) {
                    sum = cur;
                }
            }
        }
        return sum;
    }
};
class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(begin(nums), end(nums));
        int sum = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < nums.size() - 2; ++ i) {
            int j = i + 1;
            int k = nums.size() - 1;
            while (j < k) {
                int cur = nums[i] + nums[j] + nums[k];
                if (cur == target) {
                    return target;
                } else if (cur < target) {
                    j ++;
                } else {
                    k --;
                }
                if (abs(cur - target) < abs(sum - target)) {
                    sum = cur;
                }
            }
        }
        return sum;
    }
};

The optimal algorithm using two pointer algorithm is O(nlogN).

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读:
聪明的阿诺德  小学生怎样才能学好数学,成为资优生?  闹钟拨快了多少怎么算?  循环小数如何化成分数  圣经数  他是怎样猜出哪里手里的金币是单数的?  这本书共有多少页?  他们的职业各是什么?  你知道他们的名次吗?  关于逻辑推理的故事 
评论列表
添加评论