Question

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

方法一：二分查找

我们注意到数组按照非递减顺序排列，因此对于每个 $numbers[i]$，可以通过二分查找的方式找到 $target - numbers[i]$ 的位置，如果存在，那么返回 $[i, j]$ 即可。

时间复杂度 $O(n \times \log n)$，其中 $n$ 为数组 $numbers$ 的长度。空间复杂度 $O(1)$。

class Solution:
  def twoSum(self, numbers: List[int], target: int) -> List[int]:
    n = len(numbers)
    for i in range(n - 1):
      x = target - numbers[i]
      j = bisect_left(numbers, x, lo=i + 1)
      if j < n and numbers[j] == x:
        return [i, j]

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

class Solution {
public:
  vector<int> twoSum(vector<int>& numbers, int target) {
    for (int i = 0, n = numbers.size();; ++i) {
      int x = target - numbers[i];
      int j = lower_bound(numbers.begin() + i + 1, numbers.end(), x) - numbers.begin();
      if (j < n && numbers[j] == x) {
        return {i, j};
      }
    }
  }
};

func twoSum(numbers []int, target int) []int {
  for i, n := 0, len(numbers); ; i++ {
    x := target - numbers[i]
    j := sort.SearchInts(numbers[i+1:], x) + i + 1
    if j < n && numbers[j] == x {
      return []int{i, j}
    }
  }
}

function twoSum(numbers: number[], target: number): number[] {
  const n = numbers.length;
  for (let i = 0; ; ++i) {
    const x = target - numbers[i];
    let l = i + 1;
    let r = n - 1;
    while (l < r) {
      const mid = (l + r) >> 1;
      if (numbers[mid] >= x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    if (numbers[l] === x) {
      return [i, l];
    }
  }
}

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]
  }
}

方法二：双指针

我们定义两个指针 $i$ 和 $j$，分别指向数组的第一个元素和最后一个元素。每次计算 $numbers[i] + numbers[j]$，如果和等于目标值，那么返回 $[i, j]$ 即可。如果和小于目标值，那么将 $i$ 右移一位，如果和大于目标值，那么将 $j$ 左移一位。

时间复杂度 $O(n)$，其中 $n$ 为数组 $numbers$ 的长度。空间复杂度 $O(1)$。

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

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

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

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

function twoSum(numbers: number[], target: number): number[] {
  for (let i = 0, j = numbers.length - 1; ; ) {
    const x = numbers[i] + numbers[j];
    if (x === target) {
      return [i, j];
    }
    if (x < target) {
      ++i;
    } else {
      --j;
    }
  }
}

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]
  }
}