LeetCode（一）

Q&A ONE

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

public class Solution {
        public int[] twoSum(int[] nums, int target) {
                int solu[];
                for(int i=0;i<nums.length;i++){
                    for(int j=i+1;j<nums.length;j++){
                        if(nums[i]+nums[j]==target){
                           solu= new int[]{i, j};
                            return solu;
                        }
                    }
                }
            return null;
        }
    }

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Q&A TWO

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.
You may assume that each input would have exactly one solution and you may not use the same element twice.

Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

考虑到事件复杂度第一种方法已经变得不可行。由于已经排好序，所以采用一种类似于快排的思路：

从最大最小开始找（即数组的两端向中间逼进）

• 加起来太大则大的一端往中间移动
• 加起来太小则小的一端往中间移动

public int[] twoSum(int[] num, int target) {
        int[] indice = new int[2];
        if (num == null || num.length < 2) return indice;
        int left = 0, right = num.length - 1;
        while (left < right) {
            int v = num[left] + num[right];
            if (v == target) {
                indice[0] = left + 1;
                indice[1] = right + 1;
                break;
            } else if (v > target) {
                right--;
            } else {
                left++;
            }
        }
        return indice;
    }