# Leetcode 167. Two Sum II – Input array is sorted

Screenshot
from 2016-03-01 21:28:37.png

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

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.

# 1.描述

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

My code:

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.

# 2.分析

``````public class Solution { public int[] twoSum(int[] numbers, int target) { if (numbers == null || numbers.length < 2) return null; HashMap<Integer, Integer> tracker = new HashMap<Integer, Integer>(); for (int i = 0; i < numbers.length; i++) { if (tracker.containsKey(numbers[i])) { int[] ret = new int[2]; ret[0] = tracker.get(numbers[i]) + 1; ret[1] = i + 1; return ret; } else { tracker.put(target - numbers[i], i); } } return null; }}
``````
``````public int[] twoSum(int[] numbers, int target) {
int[] res = new int[2];
if (numbers == null || numbers.length < 2) {
return new int[0];
}

int left = 0, right = numbers.length - 1;
while (left < right) {
if (numbers[left] + numbers[right] == target) {
res[0] = left + 1;
res[1] = right + 1;
return res;
} else if (numbers[left] + numbers[right] > target) {
right--;
} else {
left++;
}
}

return res;
}
``````

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

# 3.代码

``````/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int* twoSum(int* numbers, int numbersSize, int target, int* returnSize) {
if (NULL == numbers || numbersSize < 2) return NULL;
unsigned int index1 = 0;
unsigned int index2 = numbersSize - 1;
while (index1 < index2) {
while (index1 < index2 && numbers[index1] + numbers[index2] < target) ++index1;
while (index1 < index2 && numbers[index1] + numbers[index2] > target) --index2;
if (numbers[index1] + numbers[index2] == target) {
*returnSize = 2;
int* returnNums = (int*)malloc((*returnSize)*sizeof(int));
returnNums[0] = index1 + 1;
returnNums[1] = index2 + 1;
return returnNums;
}
}
return NULL;
}
``````

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

Anyway, Good luck, Richardo!

# 分析

``````/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int* twoSum(int* numbers, int numbersSize, int target, int* returnSize) {
while(index1<index2)
{
if(numbers[index1]+numbers[index2]==target)
break;
else if(numbers[index1]+numbers[index2]<target)
{
index1++;
}
else
index2--;
}
int *ans=(int*)malloc(sizeof(int)*2);
ans[0]=index1+1;
ans[1]=index2+1;
*returnSize=2;
return ans;
}
``````

My code:

``````public class Solution { public int[] twoSum(int[] numbers, int target) { if (numbers == null || numbers.length == 0) { return null; } int begin = 0; int end = numbers.length - 1; while (begin < end) { int sum = numbers[begin] + numbers[end]; if (sum > target) { end--; } else if (sum < target) { begin++; } else { int[] ret = new int[2]; ret[0] = begin + 1; ret[1] = end + 1; return ret; } } return null; }}
``````

reference:

Anyway, Good luck, Richardo! — 09/02/2016