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].
Tags: Array, Hash Table
题意是让你从给定的数组中找到两个元素的和为指定值的两个索引，最容易的当然是循环两次，复杂度为O(n^2)，首次提交居然是2ms，打败了100%的提交，谜一样的结果，之后再次提交就再也没跑到过2ms了。
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};
}
}
}
return null;
}
}利用HashMap作为存储，键为目标值减去当前元素值，索引为值，比如i = 0时，此时首先要判断nums[0] = 2是否在map中，如果不存在，那么插入键值对key = 9 - 2 = 7, value = 0，之后当i = 1时，此时判断nums[1] = 7已存在于map中，那么取出该value = 0作为第一个返回值，当前i作为第二个返回值，具体代码如下所示。
class Solution {
public int[] twoSum(int[] nums, int target) {
int len = nums.length;
HashMap<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < len; ++i) {
if (map.containsKey(nums[i])) {
return new int[]{map.get(nums[i]), i};
}
map.put(target - nums[i], i);
}
return null;
}
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