题目
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].
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。 示例: 给定 nums = [2, 7, 11, 15], target = 9
因为 nums[0] + nums[1] = 2 + 7 = 9 所以返回 [0, 1]
解题思路
遍历一遍，同时使用hash把已经遍历过的值作为key，在数组中的index 存到value；遍历时将目标值减去遍历值，然后用结果去hash里找，如果有对应的值；那么直接返回就ok
解法
# ruby解法
# @param {Integer[]} nums
# @param {Integer} target
# @return {Integer[]}
def two_sum(nums, target)
search = Hash.new
nums.each_with_index do |complement, index|
i = search[target - complement]
return [i, index] if i != nil
search[complement] = index
end
raise "No two sum solution"
end
java解法
/**
* java解法
* @param {Integer[]} nums
* @param {Integer} target
* @return {Integer[]}
*/
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement)) {
return new int[] { map.get(complement), i };
}
map.put(nums[i], i);
}
throw new IllegalArgumentException("No two sum solution");
}