Two Sum - Input array is sorted

最新推荐文章于 2022-03-03 22:26:10 发布

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本文介绍了一种在已排序的整数数组中寻找两个数使其和等于特定目标值的有效算法。该算法采用双指针技术，从数组两端开始向中间逼近，直到找到满足条件的两个数或确定不存在这样的数对。

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

java

public class Solution {
    /*
     * @param nums: an array of Integer
     * @param target: target = nums[index1] + nums[index2]
     * @return: [index1 + 1, index2 + 1] (index1 < index2)
     */
    public int[] twoSum(int[] nums, int target) {
        // write your code here
        if (nums == null || nums.length == 0 || nums.length == 1) {
            return new int[]{};
        }
        int left = 0;
        int right = nums.length - 1;
        while (left <= right) {
            if (nums[left] + nums[right] > target) {
                right--;
            } else if (nums[left] + nums[right] < target) {
                left++;
            }else {
                return new int[] {left + 1, right + 1};
            }
        }
        return new int[]{-1, -1};
    }
}

python

class Solution:
    """
    @param: nums: an array of Integer
    @param: target: target = nums[index1] + nums[index2]
    @return: [index1 + 1, index2 + 1] (index1 < index2)
    """
    def twoSum(self, nums, target):
        # write your code here
        if nums is None or len(nums) == 0 or len(nums) == 1:
            return [-1, -1]
        left, right = 0, len(nums) - 1
        while left <= right:
            value = nums[left] + nums[right]
            if value > target:
                right -= 1
            elif value < target:
                left += 1
            else:
                return [left + 1, right + 1]