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 1 开始计数 ,所以答案数组应当满足 1 <= answer[0] < answer[1] <= numbers.length
。
你可以假设每个输入只对应唯一的答案,而且你不可以重复使用相同的元素。
示例 1:
输入:numbers = [2,7,11,15], target = 9 输出:[1,2] 解释:2 与 7 之和等于目标数 9 。因此 index1 = 1, index2 = 2 。
示例 2:
输入:numbers = [2,3,4], target = 6 输出:[1,3]
示例 3:
输入:numbers = [-1,0], target = -1 输出:[1,2]
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
双指针解决。
class Solution:
def twoSum(self, numbers: List[int], target: int) -> List[int]:
left, right = 0, len(numbers) - 1
while left < right:
if numbers[left] + numbers[right] == target:
return [left + 1, right + 1]
if numbers[left] + numbers[right] < target:
left += 1
else:
right -= 1
return [-1, -1]
class Solution {
public int[] twoSum(int[] numbers, int target) {
int left = 0, right = numbers.length - 1;
while (left < right) {
if (numbers[left] + numbers[right] == target) {
return new int[]{left + 1, right + 1};
}
if (numbers[left] + numbers[right] < target) {
++left;
} else {
--right;
}
}
return new int[]{-1, -1};
}
}
function twoSum(numbers: number[], target: number): number[] {
for (let right = numbers.length - 1; right >= 0; --right) {
let left = numbers.indexOf(target - numbers[right]);
if (left > -1 && left < right) {
return [left + 1, right + 1];
}
}
return [-1, -1];
}
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int left = 0, right = numbers.size() - 1;
while (left < right) {
if (numbers[left] + numbers[right] == target) {
return {left + 1, right + 1};
}
if (numbers[left] + numbers[right] < target) ++left;
else --right;
}
return {-1, -1};
}
};
func twoSum(numbers []int, target int) []int {
left, right := 0, len(numbers)-1
for left < right {
if numbers[left]+numbers[right] == target {
return []int{left + 1, right + 1}
}
if numbers[left]+numbers[right] < target {
left++
} else {
right--
}
}
return []int{-1, -1}
}