Description:
Given an array of integers, return the indices of the two numbers such that they add up to a specific target. You mat 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].
思路
首先新建一个HashMap, 把数组的元素存为key，元素的index存为value， 那么就可以利用map.containsKey()的方法找map中是否存在target-nums[i]，也就是与当前元素匹配的结果，这是判断两者的index是不是一样，不一样则可以输出，因为题目中假设只有一个解。
Solution
class solution{
public static int[] twoSum(int[] nums, int target){
Map<Integer, Integer> map = new HashMap<>();
int len = nums.length;
for(int i = 0; i < len; i++){
map.put(nums[i], i);
}
for(int j = 0; j < len; j++){
int complement = target - nums[i];
if(map.containsKey(complement) && map.get(complement) != i){
return new int[] {i, map.get(complement)}
}
}
return new IllegalArgumentException("No two sum soluton");
}
}
