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问题
给定一个整数数组和一个目标值,找出数组中和为目标值的两个数。
你可以假设每个输入只对应一种答案,且同样的元素不能被重复利用。
给定 nums = [2, 7, 11, 15], target = 9
因为 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]
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.
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
示例
public class Program {
public static void Main(string[] args) {
int[] nums = { 2, 7, 11, 15 };
var res = TwoSum(nums, 9);
var res2 = TwoSum2(nums, 9);
Console.WriteLine($"[{res[0]},{res[1]}]");
Console.WriteLine($"[{res2[0]},{res2[1]}]");
Console.ReadKey();
}
public static int[] TwoSum(int[] nums, int target) {
//类似于冒泡,双循环比较2个数的和和目标是否相等
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 };
}
}
}
//找不到时抛出异常
throw new ArgumentException("No two sum solution.");
}
public static int[] TwoSum2(int[] nums, int target) {
//用数组中的值做key,索引做value存下所有值
var dictionary = new Dictionary<int, int>();
for (int i = 0; i < nums.Length; i++) {
//记录差值
int complement = target - nums[i];
//若字典中已经存在这个值,说明匹配成功
if (dictionary.ContainsKey(complement)) {
return new int[] { dictionary[complement], i };
}
//记录索引
dictionary[nums[i]] = i;
}
//找不到时抛出异常
throw new ArgumentException("No two sum solution.");
}
}
以上给出2种算法实现,以下是这个案例的输出结果:
[0,1]
[0,1]
分析:
显而易见,TwoSum 的时间复杂度为: ,TwoSum2 的时间复杂度为: 。