[Swift]LeetCode474. 一和零 | Ones and Zeroes

In the computer world,use restricted resource you have to generate maximum benefit is what we always want to pursue.

For Now,suppose you are a dominator of m0s and n 1s respectively. On the other hand,there is an array with strings consisting of only 0s and 1s.

Now your task is to find the maximum number of strings that you can form with given m 0sand n 1s. Each 0 and 1 can be used at most once.

Note:

1. The given numbers of 0s and 1s will both not exceed 100
2. The size of given string array won‘t exceed 600. 

Example 1:

Input: Array = {"10","0001","111001","1","0"},m = 5,n = 3
Output: 4

Explanation: This are totally 4 strings can be formed by the using of 5 0s and 3 1s,which are “10,”0001”,”1”,”0” 

Example 2:

Input: Array = {"10","0","1"},m = 1,n = 1
Output: 2

Explanation: You Could form "10",but then you‘d have nothing left. Better form "0" and "1".

在计算机界中，我们总是追求用有限的资源获取最大的收益。

现在，假设你分别支配着 m 个 0 和 n 个 1。另外，还有一个仅包含 0 和 1 字符串的数组。

你的任务是使用给定的 m 个 0 和 n 个 1 ，找到能拼出存在于数组中的字符串的最大数量。每个 0 和 1 至多被使用一次。

注意:

1. 给定 0 和 1 的数量都不会超过 100。
2. 给定字符串数组的长度不会超过 600。

示例 1:

输入: Array = {"10",n = 3
输出: 4

解释: 总共 4 个字符串可以通过 5 个 0 和 3 个 1 拼出，即 "10","0" 。

示例 2:

输入: Array = {"10",n = 1
输出: 2

解释: 你可以拼出 "10"，但之后就没有剩余数字了。更好的选择是拼出 "0" 和 "1" 。

Runtime: 4936 ms

Memory Usage: 4.1 MB

 1 class Solution {
 2     func findMaxForm(_ strs: [String],_ m: Int,_ n: Int) -> Int {
 3         var dp:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:n + 1),count:m + 1)
 4         for str in strs
 5         {
 6             var zeros:Int = 0
 7             var ones:Int = 0
 8             for c in str.characters
 9             {
10                 if c == "0"
11                 {
12                     zeros += 1
13                 }
14                 else
15                 {
16                     ones += 1
17                 }
18             }
19             if zeros <= m && ones <= n
20             {
21                 for i in (zeros...m).reversed()
22                 {
23                     for j in (ones...n).reversed()
24                     {
25                         //递推公式
26                         dp[i][j] = max(dp[i][j],dp[i - zeros][j - ones] + 1)
27                     }
28                 }
29             }
30         }
31         return dp[m][n]
32     }
33 }

105934 kb

 1 class Solution {
 2     func findMaxForm(_ strs: [String],_ n: Int) -> Int {
 3         let l = strs.count
 4         var dp: [[[Int]]] = Array(repeating: Array(repeating: Array(repeating:0,count: n + 1),count: m + 1),count: l + 1)
 5         for i in 0 ... l {
 6             let counts = i == 0 ? (0,0) : getCounts(strs[i - 1])
 7             for j in 0 ... m {
 8                 for k in 0 ... n {
 9                     if i == 0 {
10                         dp[i][j][k] = 0
11                     } else if j >= counts.0 && k >= counts.1 {
12                         dp[i][j][k] = max(dp[i - 1][j][k],dp[i - 1][j - counts.0][k - counts.1] + 1)
13                     } else {
14                         dp[i][j][k] = dp[i - 1][j][k]
15                     }
16                 }
17             }
18         }
19         return dp[l][m][n]
20     }
21     
22     private func getCounts(_ str: String) -> (Int,Int) {
23         let s = Array(str)
24         var zeroCount = 0
25         s.forEach {
26             if $0 == "0" {
27                 zeroCount += 1
28             }
29         }
30         return (zeroCount,s.count - zeroCount)
31     }
32 }