第13天。

今天的题目是 Boats to Save People :

The `i`

-th person has weight `people[i]`

, and each boat can carry a maximum weight of `limit`

.

Each boat carries at most 2 people at the same time, provided the sum of the weight of those people is at most `limit`

.

Return the minimum number of boats to carry every given person. (It is guaranteed each person can be carried by a boat.)

**Example 1:**

```
Input: people = [1,2], limit = 3
Output: 1
Explanation: 1 boat (1, 2)
```

**Example 2:**

```
Input: people = [3,2,2,1], limit = 3
Output: 3
Explanation: 3 boats (1, 2), (2) and (3)
```

**Example 3:**

```
Input: people = [3,5,3,4], limit = 5
Output: 4
Explanation: 4 boats (3), (3), (4), (5)
```

**Note**:

`1 <= people.length <= 50000`

`1 <= people[i] <= limit <= 30000`

一道贪心的题目，仔细分析下题目就会发现，如果一个`weight`

比较大的人要坐船，一定是和`weight`

小的人坐船，才能保证做的船数最少。因此，只要先排序，然后在双指针判断是否能做两个人即可：

```
int numRescueBoats1(vector<int>& people, int limit) {
sort(people.begin(), people.end());
int res = 0;
int i = 0, j = people.size() -1;
while(i <= j) {
res += 1;
if (limit >= people[i] + people[j]) {
i++;
}
j--;
}
return res;
}
```