2018-01-22 TWO SUM

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) {        //满足相加为target
                indice[0] = table[target - nums[i] - min];
                indice[1] = i;
                return indice;
            }
            table[nums[i] - min] = i;
        }
    }
    free(table);
    return indice;
}

解读:

1. 第一个方法就是顺位匹配,两次循环,复杂度是O(n²);
2. 第二个方法是用匹配的方式,先计算出数组中最小的值,最大的值就是target-min;
   然后创建一个数组table,长度是max - min +1,使得 min对应数组的0位置,max对应数组的len-1位置.并初始化数组table

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

然后遍历循环nums,如果nums[ i ]位于 min --- max 之间,然后判断他对应的位置target - nums[i]-min是否有数存在,存在,则找到,否则,把table对应的位置记上标记,继续循环.
这个就是一边找,一边匹配,只是这个会耗费很大的内存~~~~.

还有一种方法,是使用哈希表,key为值,value为index,思路和上面的方法一致.利用哈希表找key的复杂度是O(1) 特性.