[Swift]LeetCode474. 一和零 | Ones and Zeroes

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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".

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在計算機界中，咱們老是追求用有限的資源獲取最大的收益。

如今，假設你分別支配着 m 個 0 和 n 個 1。另外，還有一個僅包含 0 和 1 字符串的數組。

你的任務是使用給定的 m 個 0 和 n 個 1 ，找到能拼出存在於數組中的字符串的最大數量。每一個 0 和 1 至多被使用一次。

注意:

1. 給定 0 和 1 的數量都不會超過 100。
2. 給定字符串數組的長度不會超過 600。

示例 1:

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

解釋: 總共 4 個字符串能夠經過 5 個 0 和 3 個 1 拼出，即 "10","0001","1","0" 。

示例 2:

輸入: Array = {"10", "0", "1"}, m = 1, n = 1
輸出: 2

解釋: 你能夠拼出 "10"，但以後就沒有剩餘數字了。更好的選擇是拼出 "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                         //遞推公式
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], _ m: Int, _ 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 }