problem1046

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1046. Last Stone Weight

We have a collection of rocks, each rock has a positive integer weight.

Each turn, we choose the two heaviest rocks and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is:

• If x == y, both stones are totally destroyed;
• If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.

At the end, there is at most 1 stone left. Return the weight of this stone (or 0 if there are no stones left.)

Example 1:

Input: [2,7,4,1,8,1]
Output: 1
Explanation:
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone.

Note:

1. 1 <= stones.length <= 30
2. 1 <= stones[i] <= 1000

Documentation

Index

• type PQ
  • func (pq PQ) Len() int
  • func (pq PQ) Less(i, j int) bool
  • func (pq *PQ) Pop() interface{}
  • func (pq *PQ) Push(x interface{})
  • func (pq PQ) Swap(i, j int)

Types

type PQ

type PQ []int

PQ implements heap.Interface and holds entries.

func (PQ) Len

func (pq PQ) Len() int

func (PQ) Less

func (pq PQ) Less(i, j int) bool

func (*PQ) Pop

func (pq *PQ) Pop() interface{}

Pop 从 pq 中取出最优先的 int

func (*PQ) Push

func (pq *PQ) Push(x interface{})

Push 往 pq 中放 int

func (PQ) Swap

func (pq PQ) Swap(i, j int)