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剑指 Offer II 006. 排序数组中两个数字之和

题目描述

给定一个已按照 升序排列  的整数数组 numbers ，请你从数组中找出两个数满足相加之和等于目标数 target 。

函数应该以长度为 2 的整数数组的形式返回这两个数的下标值。numbers 的下标 从 0 开始计数 ，所以答案数组应当满足 0 <= answer[0] < answer[1] < numbers.length 。

假设数组中存在且只存在一对符合条件的数字，同时一个数字不能使用两次。

 

示例 1：

输入：numbers = [1,2,4,6,10], target = 8
输出：[1,3]
解释：2 与 6 之和等于目标数 8 。因此 index1 = 1, index2 = 3 。

示例 2：

输入：numbers = [2,3,4], target = 6
输出：[0,2]

示例 3：

输入：numbers = [-1,0], target = -1
输出：[0,1]

 

提示：

• 2 <= numbers.length <= 3 * 104
• -1000 <= numbers[i] <= 1000
• numbers 按 递增顺序 排列
• -1000 <= target <= 1000
• 仅存在一个有效答案

 

注意：本题与主站 167 题相似（下标起点不同）：https://leetcode.cn/problems/two-sum-ii-input-array-is-sorted/

解法

双指针

Python3

class Solution:
    def twoSum(self, numbers: List[int], target: int) -> List[int]:
        i, j = 0, len(numbers) - 1
        while True:
            if numbers[i] + numbers[j] < target:
                i += 1
            elif numbers[i] + numbers[j] > target:
                j -= 1
            else:
                return [i, j]

Java

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

Go

func twoSum(numbers []int, target int) []int {
	for i, j := 0, len(numbers)-1; ; {
		if numbers[i]+numbers[j] < target {
			i++
		} else if numbers[i]+numbers[j] > target {
			j--
		} else {
			return []int{i, j}
		}
	}
}

C++

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int i = 0;
        int j = numbers.size() - 1;
        vector<int> res;

        while (i < j) {
            int sum = numbers[i] + numbers[j];
            if (sum < target) {
                i++;
            } else if (sum > target) {
                j--;
            } else {
                res.push_back(i);
                res.push_back(j);
                break;
            }
        }

        return res;
    }
};

Rust

use std::cmp::Ordering;

impl Solution {
    pub fn two_sum(numbers: Vec<i32>, target: i32) -> Vec<i32> {
        let n = numbers.len();
        let mut l = 0;
        let mut r = n - 1;
        loop {
            match target.cmp(&(numbers[l] + numbers[r])) {
                Ordering::Less => r -= 1,
                Ordering::Greater => l += 1,
                Ordering::Equal => break,
            }
        }
        vec![l as i32, r as i32]
    }
}

use std::cmp::Ordering;

impl Solution {
    pub fn two_sum(numbers: Vec<i32>, target: i32) -> Vec<i32> {
        let n = numbers.len();
        for i in 0..n - 1 {
            let num = target - numbers[i];
            let mut l = i + 1;
            let mut r = n - 1;
            while l <= r {
                let mid = l + (r - l) / 2;
                match num.cmp(&numbers[mid]) {
                    Ordering::Less => r = mid - 1,
                    Ordering::Greater => l = mid + 1,
                    Ordering::Equal => return vec![i as i32, mid as i32],
                }
            }
        }
        vec![-1, -1]
    }
}

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