golang力扣leetcode 2045.到达目的地的第二短时间

package main
import (
  "math"
)
type pair struct {
  nextNode, step int
}
func secondMinimum(n int, edges [][]int, time int, change int) int {
  //双向图，graph存边
  graph := make([][]int, n+1)
  for _, e := range edges {
    x, y := e[0], e[1]
    graph[x] = append(graph[x], y)
    graph[y] = append(graph[y], x)
  }
  //dist[n][0]表示1到n的最短路径，dist[n][1]表示1到n的次短路径
  dist := make([][2]int, n+1)
  for i := 1; i <= n; i++ {
    dist[i] = [2]int{math.MaxInt32, math.MaxInt32}
  }
  queue := []pair{{
    nextNode: 1,
    step:     0,
  }}
  for dist[n][1] == math.MaxInt32 {
    top := queue[0]
    queue = queue[1:]
    for _, next := range graph[top.nextNode] {
      cost := top.step + 1
      if cost < dist[next][0] {
        dist[next][0] = cost
        queue = append(queue, pair{
          nextNode: next,
          step:     cost,
        })
      } else if dist[next][0] < cost && cost < dist[next][1] {
        dist[next][1] = cost
        queue = append(queue, pair{
          nextNode: next,
          step:     cost,
        })
      }
    }
  }
  ans := 0
  for i := 1; i <= dist[n][1]; i++ {
    if ans%(2*change) >= change {
      //进入红灯区还需要等待多久才能走
      ans += 2*change - ans%(2*change)
    }
    ans += time
  }
  return ans
}