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].

int *twoSum(int *nums, int numsSize, int target) {

    int *p = (int *) malloc(sizeof(int) * 2);
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if ((nums[i] + nums[j]) == target) {
                p[0] = i;
                p[1] = j;
                return p;
            }
        }
    }
    return p;
}

int *twoSum2(int *nums, int numsSize, int target) {
    if(nums == NULL){
        return NULL;
    }
    if(numsSize < 2){
        return NULL;
    }
    int min = nums[0];
    int i = 0;
    for (i = 1; i < numsSize; i++) {
        if (nums[i] < min)
            min = nums[i];
    }
    int max = target - min;
    int len = max - min + 1;
    int *table = (int *) malloc(len * sizeof(int));
    int *indice = (int *) malloc(2 * sizeof(int));
    for (i = 0; i < len; i++) {
        table[i] = -1;         //hash初值
    }
    for (i = 0; i < numsSize; i++) {
        if (nums[i] - min < len) {
            if (table[target - nums[i] - min] != -1) {        //滿足相加為target
                indice[0] = table[target - nums[i] - min];
                indice[1] = i;
                return indice;
            }
            table[nums[i] - min] = i;
        }
    }
    free(table);
    return indice;
}

解讀:

1. 第一個(gè)方法就是順位匹配,兩次循環(huán),復(fù)雜度是O(n2);
2. 第二個(gè)方法是用匹配的方式,先計(jì)算出數(shù)組中最小的值,最大的值就是target-min;
   然后創(chuàng)建一個(gè)數(shù)組table,長(zhǎng)度是max - min +1,使得 min對(duì)應(yīng)數(shù)組的0位置,max對(duì)應(yīng)數(shù)組的len-1位置.并初始化數(shù)組table

                  len 
|----------------------|
min                    max

然后遍歷循環(huán)nums,如果nums[ i ]位于 min --- max 之間,然后判斷他對(duì)應(yīng)的位置target - nums[i]-min是否有數(shù)存在,存在,則找到,否則,把table對(duì)應(yīng)的位置記上標(biāo)記,繼續(xù)循環(huán).
這個(gè)就是一邊找,一邊匹配,只是這個(gè)會(huì)耗費(fèi)很大的內(nèi)存~~~~.

還有一種方法,是使用哈希表,key為值,value為index,思路和上面的方法一致.利用哈希表找key的復(fù)雜度是O(1) 特性.