LeetCode_每日一题今日份_167.两数之和II-输入有序数组

题解C_1

int* twoSum(int* numbers, int numbersSize, int target, int* returnSize){
    int i, j, flg = 0;

    for (i = 0; i < numbersSize - 1; i++){
        for (j = i + 1; j < numbersSize; j++){
            if (numbers[i] + numbers[j] == target){
                flg = 1;
                break;
            }
        }
        if (flg || numbers[i] > target)
            break;
    }
    *returnSize = 2;
    int *res = (int*)malloc(sizeof(int) * 2);
    res[0] = i + 1;
    res[1] = j + 1;
    return res;
}

题解C_2

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
/*
在数组中找到两个数，使得它们的和等于目标值，可以首先固定第一个数，然后寻找第二个数，第二个数等于目标值减去第一个数的差。利用数组的有序性质，可以通过二分查找的方法寻找第二个数。为了避免重复寻找，在寻找第二个数时，只在第一个数的右侧寻找。
*/
int* twoSum(int* numbers, int numbersSize, int target, int* returnSize) {
    int* ret = (int*)malloc(sizeof(int) * 2);
    *returnSize = 2;

    for (int i = 0; i < numbersSize; ++i) { //遍历数组
        int low = i + 1, high = numbersSize - 1;
        while (low <= high) {
            int mid = (high - low) / 2 + low;
            if (numbers[mid] == target - numbers[i]) {
                ret[0] = i + 1, ret[1] = mid + 1;
                return ret;
            } else if (numbers[mid] > target - numbers[i]) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
    }
    ret[0] = -1, ret[1] = -1;
    return ret;
}

模拟运行演示