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The i
-th person has weight people[i]
, and each boat can carry a maximum weight of limit
.git
Each boat carries at most 2 people at the same time, provided the sum of the weight of those people is at most limit
.github
Return the minimum number of boats to carry every given person. (It is guaranteed each person can be carried by a boat.) 微信
Example 1:ide
Input: people = [1,2], limit = 3 Output: 1 Explanation: 1 boat (1, 2)
Example 2:spa
Input: people = [3,2,2,1], limit = 3 Output: 3 Explanation: 3 boats (1, 2), (2) and (3)
Example 3:code
Input: people = [3,5,3,4], limit = 5 Output: 4 Explanation: 4 boats (3), (3), (4), (5)
Note:htm
1 <= people.length <= 50000
1 <= people[i] <= limit <= 30000
第 i
我的的体重为 people[i]
,每艘船能够承载的最大重量为 limit
。blog
每艘船最多可同时载两人,但条件是这些人的重量之和最多为 limit
。get
返回载到每个人所需的最小船数。(保证每一个人都能被船载)。
示例 1:
输入:people = [1,2], limit = 3 输出:1 解释:1 艘船载 (1, 2)
示例 2:
输入:people = [3,2,2,1], limit = 3 输出:3 解释:3 艘船分别载 (1, 2), (2) 和 (3)
示例 3:
输入:people = [3,5,3,4], limit = 5 输出:4 解释:4 艘船分别载 (3), (3), (4), (5)
提示:
1 <= people.length <= 50000
1 <= people[i] <= limit <= 30000
1 class Solution { 2 func numRescueBoats(_ people: [Int], _ limit: Int) -> Int { 3 let people = people.sorted() 4 var left = 0, right = people.count - 1 5 var count = 0 6 while left <= right { 7 if people[left] + people[right] <= limit { 8 count += 1 9 left += 1 10 right -= 1 11 } else { 12 count += 1 13 right -= 1 14 } 15 } 16 return count 17 } 18 }
1 class Solution { 2 func numRescueBoats(_ people: [Int], _ limit: Int) -> Int { 3 var people = people.sorted(by:<) 4 var ans:Int = 0 5 var hi:Int = people.count - 1 6 var lo:Int = 0 7 while (hi >= lo) 8 { 9 if people[lo] + people[hi] <= limit 10 { 11 lo += 1 12 } 13 hi -= 1 14 ans += 1 15 } 16 return ans 17 } 18 }
764ms
1 class Solution { 2 func numRescueBoats(_ people: [Int], _ limit: Int) -> Int { 3 var people = people.sorted() 4 var ans = 0, headIndex = 0, tailIndex = people.count - 1 5 while headIndex <= tailIndex { 6 ans += 1 7 var lastTailIndex = tailIndex 8 while tailIndex > headIndex { 9 if people[tailIndex]+people[headIndex] > limit { 10 tailIndex -= 1 11 if tailIndex == headIndex { 12 ans += (lastTailIndex - headIndex) 13 return ans 14 } 15 }else { 16 ans += (lastTailIndex - tailIndex) 17 tailIndex -= 1 18 break 19 } 20 } 21 headIndex += 1 22 } 23 return ans 24 } 25 }