定义five,ten为顾客付5元和10元的张数，找零时首先找较大面额，分情况讨论：
如果顾客付5元，直接five++
如果顾客付10元，自己至少有一张5元，five–,ten++;否则返回false
如果顾客付20元，自己至少有一张5元和一张10元，five–,ten–；或者自己至少有3张5元，five-=3；否则返回false

func lemonadeChange(bills []int) bool {five, ten := 0, 0for i := 0; i < len(bills); i++ {if bills[i] == 5 {five++} else if bills[i] == 10 {if five > 0 {ten++five--} else {return false}} else {if ten > 0 && five > 0 {ten--five--} else if five > 2 {five -= 3} else {return false}}}return true
}

先对数组中数对的第一个元素降序排序，然后对数对的第二个元素插入升序排序

func reconstructQueue(people [][]int) [][]int {// 1.高个子对矮个子没影响，先降序排序sort.Slice(people, func(i, j int) bool {// 身高相同时，序号小在前if people[i][0] == people[j][0] {return people[i][1] < people[j][1]}return people[i][0] > people[j][0]})// 2.然后把矮个子插入对应位置person[i][1]即可res := make([][]int, len(people))for i := 0; i < len(people); i++ {if len(res) <= people[i][1] {res = append(res, people[i])} else {// 在指定位置插入元素index := people[i][1]copy(res[index+1:], res[index:])res[index] = people[i]}}return res
}

原地排序

for index, person := range people {copy(people[person[1]+1:index+1], people[person[1]:index+1])people[person[1]] = person
}

贪心：对气球右边界排序，初始时最大右边界为points[0][1]，后续遍历的气球右边界都比初始值大，只要保证气球左边界小于maxRight就能被箭射穿，否则更新最大右边界，箭个数++

func findMinArrowShots(points [][]int) int {sort.Slice(points, func(i, j int) bool {return points[i][1] < points[j][1]})res := 1maxRight := points[0][1]for i := 1; i < len(points); i++ {if points[i][0] <= maxRight {continue}maxRight = points[i][1]res++}return res
}