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Commit96fe5ee

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Merge pull request6boris#445 from 0xff-dev/master
Add solution and test-cases for problem 502
2 parents7a0400f +4ed22b6 commit96fe5ee

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3 files changed

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‎leetcode/501-600/0502.IPO/README.md

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Original file line numberDiff line numberDiff line change
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#[502.IPO][title]
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>[!WARNING|style:flat]
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>This question is temporarily unanswered if you have good ideas. Welcome to[Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
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##Description
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Suppose LeetCode will start its**IPO** soon. In order to sell a good price of its shares to Venture Capital, LeetCode would like to work on some projects to increase its capital before the**IPO**. Since it has limited resources, it can only finish at most`k` distinct projects before the**IPO**. Help LeetCode design the best way to maximize its total capital after finishing at most`k` distinct projects.
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You are given`n` projects where the i<sup>th</sup> project has a pure profit`profits[i]` and a minimum capital of`capital[i]` is needed to start it.
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Initially, you have`w` capital. When you finish a project, you will obtain its pure profit and the profit will be added to your total capital.
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Pick a list of**at most**`k` distinct projects from given projects to**maximize your final capital**, and return the final maximized capital.
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The answer is guaranteed to fit in a 32-bit signed integer.
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**Example 1:**
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```
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Input: a = "11", b = "1"
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Output: "100"
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Input: k = 2, w = 0, profits = [1,2,3], capital = [0,1,1]
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Output: 4
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Explanation: Since your initial capital is 0, you can only start the project indexed 0.
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After finishing it you will obtain profit 1 and your capital becomes 1.
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With capital 1, you can either start the project indexed 1 or the project indexed 2.
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Since you can choose at most 2 projects, you need to finish the project indexed 2 to get the maximum capital.
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Therefore, output the final maximized capital, which is 0 + 1 + 3 = 4.
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```
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##题意
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>...
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##题解
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**Example 2:**
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###思路1
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>...
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IPO
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```go
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```
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Input: k = 3, w = 0, profits = [1,2,3], capital = [0,1,2]
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Output: 6
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```
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##结语
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‎leetcode/501-600/0502.IPO/Solution.go

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Original file line numberDiff line numberDiff line change
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package Solution
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funcSolution(xbool)bool {
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import (
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"container/heap"
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"sort"
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)
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typeheap502Itemstruct {
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cost,profitint
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}
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typeheap502ItemList []heap502Item
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func (h*heap502ItemList)Len()int {
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returnlen(*h)
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}
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func (h*heap502ItemList)Less(i,jint)bool {
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a,b:= (*h)[i], (*h)[j]
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ifa.profit==b.profit {
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returna.cost<b.cost
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}
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returna.profit>b.profit
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}
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func (h*heap502ItemList)Swap(i,jint) {
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(*h)[i], (*h)[j]= (*h)[j], (*h)[i]
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}
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func (h*heap502ItemList)Push(xinterface{}) {
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*h=append(*h,x.(heap502Item))
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}
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func (h*heap502ItemList)Pop()interface{} {
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old:=*h
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l:=len(old)
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x:=old[l-1]
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*h=old[:l-1]
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returnx
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}
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funcSolution(kint,wint,profits []int,capital []int)int {
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x:=make([]heap502Item,len(profits))
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fori:=0;i<len(profits);i++ {
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x[i]=heap502Item{cost:capital[i],profit:profits[i]}
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}
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sort.Slice(x,func(i,jint)bool {
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returnx[i].cost<x[j].cost
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})
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h:=heap502ItemList{}
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pick:=0
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for ;k>0;k-- {
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for ;pick<len(x)&&x[pick].cost<=w;pick++ {
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heap.Push(&h,x[pick])
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}
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iflen(h)==0 {
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break
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}
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x:=heap.Pop(&h).(heap502Item)
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w+=x.profit
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}
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returnw
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}

‎leetcode/501-600/0502.IPO/Solution_test.go

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Original file line numberDiff line numberDiff line change
@@ -10,30 +10,30 @@ func TestSolution(t *testing.T) {
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//测试用例
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cases:= []struct {
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namestring
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inputsbool
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expectbool
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k,wint
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p,c []int
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expectint
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}{
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{"TestCase",true,true},
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{"TestCase",true,true},
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{"TestCase",false,false},
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{"TestCase1",2,0, []int{1,2,3}, []int{0,1,1},4},
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{"TestCase2",3,0, []int{1,2,3}, []int{0,1,2},6},
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}
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//开始测试
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fori,c:=rangecases {
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t.Run(c.name+" "+strconv.Itoa(i),func(t*testing.T) {
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got:=Solution(c.inputs)
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got:=Solution(c.k,c.w,c.p,c.c)
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if!reflect.DeepEqual(got,c.expect) {
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t.Fatalf("expected: %v, but got: %v, with inputs: %v",
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c.expect,got,c.inputs)
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t.Fatalf("expected: %v, but got: %v, with inputs: %v %v %v %v",
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c.expect,got,c.k,c.w,c.p,c.c)
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}
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})
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}
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}
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//压力测试
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//压力测试
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funcBenchmarkSolution(b*testing.B) {
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}
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//使用案列
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//使用案列
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funcExampleSolution() {
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}

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