版权声明:原创勿转 https://cloud.tencent.com/developer/article/1420469
思路
按照面额不同的情况分别考虑
code
代码语言:javascript复制func lemonadeChange(bills []int) bool {
m := make([]int, 2)
for i := 0; i < len(bills); i {
switch bills[i] {
case 5:
m[0]
break
case 10:
if m[0] <= 0 {
return false
}
m[0]--
m[1]
break
case 20:
flag := false
if m[0] >= 1 && m[1] >= 1 {
flag = true
m[0]--
m[1]--
} else if m[0] >= 3 {
flag = true
m[0] -= 3
}
if !flag {
return false
}
break
}
}
return true
}