这篇博客详细介绍了如何使用哈希表解决LeetCode上的两数之和问题。作者提供了两种解决方案，一种是Java实现，另一种是C++实现。在Java方案中，使用了HashMap存储数组元素的值及其索引，通过遍历数组，查找目标值减去当前值是否存在哈希表中，从而找到两个数的索引。C++方案同样利用了哈希表，但采用了双指针的方法，避免了二次遍历。这两种方法都能够有效地找到数组中和为目标值的两个数的索引。
Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Example 1:
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].
Example 2:
Input: nums = [3,2,4], target = 6
Output: [1,2]
Example 3:
Input: nums = [3,3], target = 6
Output: [0,1]
Constraints:
2 <= nums.length <= 103
-109 <= nums[i] <= 109
-109 <= target <= 109
Only one valid answer exists.
Accepted
3.6M
Submissions
7.8M
class Solution {
public int[] twoSum(int[] nums, int target) {
int[] indices = new int[2];
HashMap tmpmap = new HashMap(); // key为nums的值，value为nums下标
for (int i = 0; i < nums.length; ++i){
if(tmpmap.containsKey(Integer.valueOf(nums[i]))){
indices[0] = tmpmap.get(Integer.valueOf(nums[i])).intValue();
indices[1] = i;
break;
}
tmpmap.put(Integer.valueOf(target-nums[i]), Integer.valueOf(i));
}
return indices;
}
}
class Solution {
public:
/*vector twoSum(vector& nums, int target) {
vector indices(2);
int len = nums.size();
for(int i = 0; i < len; ++i){
for(int j = i+1; j < len; ++j){
if(nums[i] + nums[j] == target){
indices[0] = i;
indices[1] = j;
break;
}
}
}
return indices;
}*/
vector twoSum(vector& nums, int target){
vector twoSum;
map tmpmap;//键值为nums的值，变量值为nums下标
for(int i=0;i
if(tmpmap.count(nums[i])!=0){
twoSum.push_back(tmpmap[nums[i]]);
twoSum.push_back(i);
break;
}
tmpmap[target-nums[i]]=i;
}
return twoSum;
}
};
标签：Java,target,nums,int,Sum,tmpmap,twoSum,indices,LeetCode
来源： https://blog.youkuaiyun.com/itorly/article/details/110294000
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