import java.util.HashMap;
public class TwoSum {
//测试
public static void main(String[] args) {
int[] nums = {2,7,9,11};
int target = 9;
TwoSum twoSum = new TwoSum();
int[] ints = twoSum.twoSum3(nums, target);
for (int i = 0; i <ints.length ; i++) {
System.out.println(ints[i]);
}
}
/**
* 方法一:暴力破解 时间复杂度O(n^2) 空间复杂度O(1)
* @param nums
* @param target
* @return
*/
public int[] twoSum(int[] nums, int target) {
int n = nums.length;
for (int i = 0; i < n-1; i++) {
for (int j = i+1; j < n; j++) {
if(nums[i]+nums[j]==target){
return new int[]{i,j};
}
}
}
//如果找不到 抛出异常
throw new IllegalArgumentException("no solution");
}
/**
* 方法二:哈希表保存所有数的信息 时间复杂度O(n) 空间复杂度O(n)
* @param nums
* @param target
* @return
*/
public int[] twoSum2(int[] nums, int target) {
int n = nums.length;
//定义一个哈希表
HashMap<Integer,Integer> map = new HashMap<>();
//1.遍历数组,把所有数据保存到hash表
for (int i = 0; i < n; i++) {
map.put(nums[i],i);
}
//2.再次遍历数组,寻找每个数对应的哪个数是否存在
for (int i = 0; i < n; i++) {
int thatNum = target-nums[i];
//如果哪个数存在,并且不是当前数自身,就直接返回结果
if(map.containsKey(thatNum)&&map.get(thatNum)!=i){
return new int[]{i,map.get(thatNum)};
}
}
//如果找不到 抛出异常
throw new IllegalArgumentException("no solution");
}
/**
* 方法三:遍历一次哈希表 时间复杂度O(n) 空间复杂度O(n)
* @param nums
* @param target
* @return
*/
public int[] twoSum3(int[] nums, int target) {
int n = nums.length;
//定义一个哈希表
HashMap<Integer,Integer> map = new HashMap<>();
//遍历数组,寻找每个数对应的哪个数是否存在
for (int i = 0; i < n; i++) {
int thatNum = target-nums[i];
//如果哪个数存在,并且不是当前数自身,就直接返回结果
if(map.containsKey(thatNum)&&map.get(thatNum)!=i){
return new int[]{map.get(thatNum),i};
}
map.put(nums[i],i);
}
//如果找不到 抛出异常
throw new IllegalArgumentException("no solution");
}
}