> For the complete documentation index, see [llms.txt](https://hadyang.gitbook.io/interview/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://hadyang.gitbook.io/interview/content/docs/offer/findnumberswithsum/index.md).

# index

***

### title: 和为 date: 2019-08-21T11:00:41+08:00 draft: false categories: offer

### 题目

[牛客网](https://www.nowcoder.com/practice/390da4f7a00f44bea7c2f3d19491311b?tpId=13\&tqId=11195\&tPage=3\&rp=2\&ru=%2Fta%2Fcoding-interviews\&qru=%2Fta%2Fcoding-interviews%2Fquestion-ranking)

输入一个递增排序的数组和一个数字S，在数组中查找两个数，使得他们的和正好是S，如果有多对数字的和等于S，输出两个数的乘积最小的。

> 对应每个测试案例，输出两个数，小的先输出。

### 解题思路

1. 利用二分查找的思想，由于是排序数组，通过两个指针来进行遍历

```
public ArrayList<Integer> FindNumbersWithSum(int[] array, int sum) {
    ArrayList<Integer> res = new ArrayList<>();

    if (array == null || array.length == 1) {
        return res;
    }

    int start = 0, end = array.length - 1;
    int minMulti = Integer.MAX_VALUE;
    int a = -1, b = -1;

    while (start < end) {
        int t = array[start] + array[end];
        if (t == sum) {
            int multi = array[start] * array[end];
            if (multi < minMulti) {
                a = array[start];
                b = array[end];
                minMulti = multi;
            }
            start++;
            end--;
        } else if (t > sum) end--;
        else start++;
    }

    if (a == -1 || b == -1) {
        return res;
    }

    res.add(a);
    res.add(b);

    return res;
}
```


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