LeetCode (Easy Part) Two Sum
2018-04-05 19:22
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No1. TwoSum
题目:Given an array of integers, return indices of the two numbers such that they add up to a specific target.给定一个整数数组,返回这两个数字的索引,使它们合计成一个特定的目标。
You may assume that each input would have exactly one solution, and you may not use the same element twice.您可能会认为每个输入都只有一个解决方案,并且您可能不会使用相同的元素两次。Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].思路:
定义两个索引值 i,j 均代表数组的下标。从左到右依次遍历整个数组 num[i],num[j]为任何有可能的两个数的组合,直到两数之和等于先前给的target值,返回此时所对应的数组下标。
代码实现: public class Solution
{
public int[] TwoSum(int[] nums, int target)
{
for (int i = 0; i < nums.Length; i++)
{
for (int j = i + 1; j < nums.Length; j++)
{
if (nums[i] + nums[j] == target)
{
return new int[]{i,j};
}
}
}
throw new Exception("error");
}
}
题目:Given an array of integers, return indices of the two numbers such that they add up to a specific target.给定一个整数数组,返回这两个数字的索引,使它们合计成一个特定的目标。
You may assume that each input would have exactly one solution, and you may not use the same element twice.您可能会认为每个输入都只有一个解决方案,并且您可能不会使用相同的元素两次。Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].思路:
定义两个索引值 i,j 均代表数组的下标。从左到右依次遍历整个数组 num[i],num[j]为任何有可能的两个数的组合,直到两数之和等于先前给的target值,返回此时所对应的数组下标。
代码实现: public class Solution
{
public int[] TwoSum(int[] nums, int target)
{
for (int i = 0; i < nums.Length; i++)
{
for (int j = i + 1; j < nums.Length; j++)
{
if (nums[i] + nums[j] == target)
{
return new int[]{i,j};
}
}
}
throw new Exception("error");
}
}
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