Two Sum and Two Sum II

给定一个整形的数组，找出其中的两个数使其和为某个指定的值，并返回这两个数的下标，假设数组元素的值各不相同

public class TwoSum {

	//时间复杂度为o(n)
	public int[] twoSum2(int[] nums, int target) {
		int [] res ={-1,-1};
     	Map map = new HashMap();
     	for (int i = 0; i < nums.length; i++) {
			map.put(nums[i], i);
		}
     	for (int j = 0; j < nums.length; j++) {
			if (map.containsKey(target-nums[j]) && target != 2*nums[j]) {
				res[0] = j;
				res[1] = map.get(target-nums[j]);
				break;
			}
		}
	return res;
 }
}

Two Sum II （有序数组，有重复元素）

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.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int[] res = {-1,-1};
	    int begin = 0;
	    int end = numbers.length-1;
	    while (begin < end) {
			if (numbers[begin] + numbers[end] == target) {
				res[0] = begin + 1;
				res[1] = end +1;
				break;
			}else if (numbers[begin] + numbers[end] > target) {
				end--;
			}else {
				begin++;
			}
		}
	    return res;
    }
}