LeetCode 1.两数之和 暴力破解和采用HashMap解法

**

**
分享三种解法。

1.暴力破解，双重for循环遍历nums数组，找到对应的两个下标。算法复杂度为O(n^2)

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[i] + nums[j] == target) {
                    return new int[]{i, j};
                }
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

2.采用HashMap，存储数组元素和数组元素的下标，利用键值关系来解题。

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            map.put(nums[i],i);
        }
        for (int j = 0; j < nums.length; j++) {
            int complement = target - nums[j];
            if (map.containsKey(complement) && map.get(complement) != j) {
                return new int[] {j , map.get(complement)};
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

3.对采用HashMap的解法继续优化，由于一个元素的下标不能重复出现在答案中，故可以边找边往map中放键值对，数组的元素都会基于前面的元素来查找是否符合条件。

class Solution {
    public int[] twoSum(int[] nums, int target) {
        HashMap map = new HashMap<>();
        for (int j = 0 ; j < nums.length; j++) {
            int complement = target - nums[j];
            if (map.containsKey(complement)) {
                return new int[]{map.get(complement),j};
            }
            map.put(nums[j], j);
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}