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  • Leetcode167-Two Sum II Input array is sorted-Easy

    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.

    Note:

    • Your returned answers (both index1 and index2) are not zero-based.
    • You may assume that each input would have exactly one solution and you may not use the same element twice.

    Example:

    Input: numbers = [2,7,11,15], target = 9
    Output: [1,2]
    Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
     
    思路:
    two pointers法:i,j 分别指向sorted array的头和尾,如果numbers[i] + numbers[j] < target, 左指针后移,否则有指针左移。当i = j时,则没有找到和为target对应的index。则返回空。
     
    注意:
    j = numbers.length - 1; 不要再忘了-1 才是数组最后一位的index!!
     
    代码:
        public int[] twoSum(int[] numbers, int target) {
            
            int i = 0, j = numbers.length - 1;
            
            while (i != j) {
                int sum = numbers[i] + numbers[j];
                if (sum == target) {
                    return new int[]{i+1, j+1};
                }
                else if (sum > target) {
                    j--;
                }
                else {
                    i++;
                }       
        }
            return new int[]{};
    }
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  • 原文地址:https://www.cnblogs.com/Jessiezyr/p/10628059.html
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