leetcode题目之两数求和

题目很简单，返回数组中等于目标的两数的下标。用了两种方法去求解。第一种两层循环，时间复杂度O（n^2） 第二种用了hashmap。时间复杂度O（n）.

question:

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].

解题 法1：

public static 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[j] == target - nums[i]) {
return new int[] { i, j };
}
}
}
throw new IllegalArgumentException("No two sum solution");
}

法2：

public static int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
map.put(nums[i], i);
}
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement) && map.get(complement) != i) {
return new int[] { i, map.get(complement) };
}
}
throw new IllegalArgumentException("No two sum solution");
}

测试：

public static void main(String[] args)
{
int[] nums={1,0,4,8};
twoSum(nums,9);
int[] num1=twoSum(nums,9);
System.out.println(Arrays.toString(num1));

}

输出：[0,3]