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].
初始代码
利用了两个指针。
class Solution {
public int[] twoSum(int[] nums, int target) {
ArrayList<Integer> list = new ArrayList<Integer>();
ArrayList<Integer> unSortedlist = new ArrayList<Integer>();
for(int i:nums){
list.add(i);
unSortedlist.add(i);
}
Collections.sort(list);
int first = 0;
int last = list.size()-1;
while(last>first){
if(list.get(first)+list.get(last)<target){
first++;
}else if(list.get(first)+list.get(last)>target){
last--;
}else{
break;
}
}
ArrayList<Integer> valueTwoList = new ArrayList<Integer>();
for(int i = 0;i<nums.length;i++){
if(nums[i] ==list.get(first)||nums[i] ==list.get(last)){
valueTwoList.add(i);
break;
}
}
for(int i = nums.length-1;i>=0;i--){
if(nums[i] ==list.get(first)||nums[i] ==list.get(last)){
valueTwoList.add(i);
break;
}
}
Collections.sort(valueTwoList);
int result[] = {valueTwoList.get(0),valueTwoList.get(1)};
return result;
}
}
大神代码 利用了containskey ,相当厉害了。
public int[] twoSum(int[] nums, int target) {
int[] result = new int[2];
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i = 0; i < nums.length; i++) {
if (map.containsKey(target - nums[i])) {
result[1] = i;
result[0] = map.get(target - nums[i]);
return result;
}
map.put(nums[i], i );
}
return result;
}