#1_两数之和

Category	Difficulty	Likes	Dislikes
algorithms	Easy (47.92%)	7893	-

我的答案

public int[] twoSum(int[] nums, int target) {
    int i, j;
    int[] res = new int[2];
    for(i = 0; i < nums.length; i++){
        for(j = i + 1; j < nums.length; j++){
            if(nums[i] + nums[j] == target){
                res[0] = i;
                res[1] = j;
                return res;
            }
        }
    }
    return res;
}

解题思路

遍历每个元素，并查找是否有一个元素与它相加等于 target

答案分析

时间复杂度 O(n²)

空间复杂度 O(1)

缺少异常处理

参考方案

public int[] twoSum(int[] nums, int target) {
    Map<Integer,Integer> map = new HashMap<>();
    for(int i = 0; i < nums.length; i++){
        int tmp = target - nums[i];
        if(map.containsKey(tmp)){
            return new int[] {map.get(tmp),i}; 
        }
        map.put(nums[i], i);
    }
    throw new IllegalArgumentException("No Two Sum Solution");
}

时间复杂度 O(n)

空间复杂度 O(n)

备注

• HashMap 的 API

void                 clear()
Object               clone()
boolean              containsKey(Object key)
boolean              containsValue(Object value)
Set<Entry<K, V>>     entrySet()
V                    get(Object key)
boolean              isEmpty()
Set<K>               keySet()
V                    put(K key, V value)
void                 putAll(Map<? extends K, ? extends V> map)
V                    remove(Object key)
int                  size()
Collection<V>        values()