- Method1: Search
We need to find if (target - current) is in the set, if not, add current to a hashSet for checking.
Since to need to return the index according to value, so the key is value of number and value is index of number in array.
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
int numLength = nums.length;
for(int i = 0; i < numLength; i++){
if(!map.containsKey(target - nums[i])){
map.put(nums[i], i);
}else{
return new int[]{i, map.get(target - nums[i])};
}
}
return null;
}
}
- 二刷的时候仍然想到用hash表解决该问题。键存的是数字的值,值存的是index。
- 在遍历的时候,我们同时检查target - nums[i]是否存在。
- 如果不存在,我们将数值加入哈希表。
- 如果存在,我们直接返回index。
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for(int i = 0; i < nums.length; i++){
if(!map.containsKey(target - nums[i]))
map.put(nums[i], i);
else
return new int[]{i, map.get(target - nums[i])};
}
return null;
}
}