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 [

這道題的思路是兩個迴圈遍歷找相加等於target的數。程式碼如下：

int* twoSum(int* nums, int numsSize, int target) {
    int res[2];
    int i;
    for(i=0;i<target;i++)
        for(int j=i+1;j<target;j++)
    {
        if(target==nums[i]+nums[j])
        {
            res[0]=i;
            res[1]=j;
            break;
        }
    }
    return res;
}
 

我嘗試下用malloc進行動態分配記憶體到時無補於事，於是我用其他辦法：用java方法就可以搞定了，程式碼如下：

public class Solution {
  public static int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        if (nums == null || nums.length == 0) {
            return result;
        }

        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] + nums[j] == target) {
                    result[0] = i;
                    result[1] = j;
                    return result;
                }
            }
        }
        return result;
    }
}