LeetCode_1_TwoSum

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2019 年 03 月 03 日

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1927字数

1. 题目描述

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].

2. 初解

遍历两层数组，一个一个加直到找出target

时间复杂度O(n^2)

int* twoSum(int* nums, int numsSize, int target) {
    int* arr = NULL;
    arr = (int* ) malloc(8);
    int i = 0, j = 0;

    for(; i < numsSize; i++) {
        for(j=i+1; j < numsSize; j++) {
            if(*(nums+i) + *(nums+j) == target) {
            *arr = i;
            *(arr+1) = j;
            return arr;
            }
        }
    }
    return NULL;
}

3. 根据网上思路优化解法

用线性探测法的哈希表解决本题。
遍历一遍给定数组，当遍历到某个元素时，如果这个元素与某个已经加入到哈希表中的元素相加等于target，则直接返回下标，否则将该元素加入哈希表。

时间复杂度O(n)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define HTCAPACITY 50000

int hash(int key) {
        int ret = key % HTCAPACITY;
        return (ret < 0)?(ret + HTCAPACITY):(ret);
}

void htInsert(int *keys, int *values, int key, int value) {
    int index = hash(key);

    while(values[index] >= 0)
        index = (index + 1) % HTCAPACITY;
    keys[index] = key;
    values[index] = value;
}

int htSearch(int *keys, int *values, int key) {
    int index = hash(key);

    while(values[index] >= 0) {
        if(keys[index] == key)
            return values[index];
        index = (index + 1) % HTCAPACITY;
    }
    return -1;
}

// keys[index]和values[index]共同描述索引为index的哈希表项
// keys[index]表示该项的key
// values[index]为-1时表示该索引下无项
// values[index]为大于-1的整数时表示该项对应的key在题目给定数组nums[]中的下标
int* twoSum(int* nums, int numsSize, int target) {
    int keys[HTCAPACITY] = {0};
    int values[HTCAPACITY];
    memset(values, -1, sizeof(int)*HTCAPACITY);
    int *indices = (int *)malloc(sizeof(int)*2);
    int value = -1, i = 0, complement;

    for(i = 0; i<numsSize; i++) {
        complement = target - nums[i];
        if((value = htSearch(keys, values, complement)) != -1) {
            indices[0] = value;
            indices[1] = i;
            return indices;
        } else {
            htInsert(keys, values, nums[i], i);
            print(keys, values);
        }
    }
    return NULL;
}

int main(int argc, char const *argv[])
{
    int nums[] = {0, 1, 2, 0};
    int *res = twoSum(nums, 4, 0);
    printf("%d %d\n", res[0], res[1]);
    return 0;
}

更新 C++ 版

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        map<int, vector<int>::size_type> res;
                                                                                                                                        
        for(unsigned i = 0; i < nums.size(); i++) {
            int diff = target - nums[i];

            if(res.find(diff) != res.end() && res[diff] != i) {
                return {(int)res[diff], (int)i};
            }
            else {
                res[nums[i]] = i;
            }
        }

        return {};
    }
};