解题思路一;
1.数组nums, 和目标值target
2.遍历数组元素a,是否存在一个值= target-a;
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[j]==target-nums[i]) {
return new int[]{i,j};
}
}
}
throw new IllegalArgumentException("no");
}
}
解题思路二
1.创建一个HashMap
Map<Integer, Integer> map = new HashMap<>();
2.遍历数组,将数组中的元素放入map中
map.put(nums[i], i);
3.遍历map,找到是否存在一个值= target-a;
int complement = target - nums[i];
4,存在返回
return new int[] { i, map.get(complement) };
完整代码
class Solution {
public int[] twoSum(int[] nums,int target) {
Map<Integer,Integer>map = new HashMap<>();
for(int i =0;i<nums.length;i++) {
map.put(nums[i],i);
}
for(int i =0;i<nums.length;i++) {
int complement= target -nums[i];
if(map.containsKey(complement)&&map.get(complement)!=i){
return new int[] {i,map.get(complement)};
}
}
throw new IllegalArgumentException("No two sum solution");
}
}