leetcode 167. 两数之和 II - 输入有序数组 从暴力解法到二分法到左右指针

原题链接

167. 两数之和 II - 输入有序数组

暴力解法

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int n = numbers.length;
        for (int i = 0; i < n; i++) {
            for (int j = i+1; j < n; j++) {
                if (numbers[i] + numbers[j] == target) return new int[]{i+1,j+1};
            }
        }
        return new int[0];
    }
}

• 时间复杂度为 O(N²)
• 空间复杂度为 O(1)

二分法改进

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int n = numbers.length;
        for (int i = 0; i < n; i++) {
            int l = i + 1, r = n-1, mid;
            while (l <= r) {
                mid = (l + r) / 2;
                if (numbers[mid] == target - numbers[i])
                    return new int[] { i + 1, mid + 1 };
                else if (numbers[mid] < target - numbers[i])
                    l = mid + 1;
                else
                    r = mid - 1;
            }
        }
        return new int[0];
    }
}

• 时间复杂度为 O(NlogN)
• 空间复杂度为 O(1)

左右指针

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int n = numbers.length, l = 0, r = n - 1, sum;
        while (l < r) {
            sum = numbers[l] + numbers[r];
            if (sum == target)
                return new int[] { l + 1, r + 1 };
            if (sum < target)
                l++;
            if (sum > target)
                r--;
        }
        return new int[0];
    }
}

• 时间复杂度为 O(N)
• 空间复杂度为 O(1)