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Commit 4f89e08

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feat: add solutions to lc problem: No.1363 (doocs#1637)
No.1363.Largest Multiple of Three
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‎solution/1300-1399/1363.Largest Multiple of Three/README.md‎

Lines changed: 184 additions & 1 deletion
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:贪心 + 动态规划 + 逆推**
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我们定义 $f[i][j]$ 表示在前 $i$ 个数中选取若干个数,使得选取的数的和模 3ドル$ 为 $j$ 的最大长度。为了使得选取的数最大,我们需要尽可能选取更多的数,因此我们需要使得 $f[i][j]$ 尽可能大。我们初始化 $f[0][0] = 0,ドル其余的 $f[0][j] = -\infty$。
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考虑 $f[i][j]$ 如何进行状态转移。我们可以不选取第 $i$ 个数,此时 $f[i][j] = f[i - 1][j]$;我们也可以选取第 $i$ 个数,此时 $f[i][j] = f[i - 1][(j - x_i \bmod 3 + 3) \bmod 3] + 1,ドル其中 $x_i$ 表示第 $i$ 个数的值。因此我们有如下的状态转移方程:
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$$
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f[i][j] = \max \{ f[i - 1][j], f[i - 1][(j - x_i \bmod 3 + 3) \bmod 3] + 1 \}
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$$
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如果 $f[n][0] \le 0,ドル那么我们无法选取任何数,因此答案字符串为空。否则我们可以通过 $f$ 数组进行逆推,找出选取的数。
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定义 $i = n,ドル $j = 0,ドル从 $f[i][j]$ 开始逆推,记 $k = (j - x_i \bmod 3 + 3) \bmod 3,ドル如果 $f[i - 1][k] + 1 = f[i][j],ドル那么我们选取了第 $i$ 个数,否则我们没有选取第 $i$ 个数。如果我们选取了第 $i$ 个数,那么我们将 $j$ 更新为 $k,ドル否则我们保持 $j$ 不变。为了使得同等长度的数最大,我们应该优先选取较大的数,因此,我们在前面首先对数组进行排序。
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时间复杂度 $O(n \times \log n),ドル空间复杂度 $O(n)$。其中 $n$ 为数组的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def largestMultipleOfThree(self, digits: List[int]) -> str:
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digits.sort()
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n = len(digits)
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f = [[-inf] * 3 for _ in range(n + 1)]
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f[0][0] = 0
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for i, x in enumerate(digits, 1):
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for j in range(3):
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f[i][j] = max(f[i - 1][j], f[i - 1][(j - x % 3 + 3) % 3] + 1)
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if f[n][0] <= 0:
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return ""
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arr = []
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j = 0
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for i in range(n, 0, -1):
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k = (j - digits[i - 1] % 3 + 3) % 3
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if f[i - 1][k] + 1 == f[i][j]:
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arr.append(digits[i - 1])
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j = k
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i = 0
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while i < len(arr) - 1 and arr[i] == 0:
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i += 1
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return "".join(map(str, arr[i:]))
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String largestMultipleOfThree(int[] digits) {
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Arrays.sort(digits);
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int n = digits.length;
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int[][] f = new int[n + 1][3];
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final int inf = 1 << 30;
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for (var g : f) {
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Arrays.fill(g, -inf);
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}
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f[0][0] = 0;
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for (int i = 1; i <= n; ++i) {
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for (int j = 0; j < 3; ++j) {
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f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - digits[i - 1] % 3 + 3) % 3] + 1);
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}
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}
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if (f[n][0] <= 0) {
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return "";
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}
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StringBuilder sb = new StringBuilder();
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for (int i = n, j = 0; i > 0; --i) {
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int k = (j - digits[i - 1] % 3 + 3) % 3;
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if (f[i - 1][k] + 1 == f[i][j]) {
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sb.append(digits[i - 1]);
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j = k;
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}
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}
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int i = 0;
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while (i < sb.length() - 1 && sb.charAt(i) == '0') {
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++i;
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}
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return sb.substring(i);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string largestMultipleOfThree(vector<int>& digits) {
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sort(digits.begin(), digits.end());
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int n = digits.size();
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int f[n + 1][3];
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memset(f, -0x3f, sizeof(f));
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f[0][0] = 0;
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for (int i = 1; i <= n; ++i) {
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for (int j = 0; j < 3; ++j) {
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f[i][j] = max(f[i - 1][j], f[i - 1][(j - digits[i - 1] % 3 + 3) % 3] + 1);
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}
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}
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if (f[n][0] <= 0) {
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return "";
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}
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string ans;
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for (int i = n, j = 0; i; --i) {
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int k = (j - digits[i - 1] % 3 + 3) % 3;
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if (f[i - 1][k] + 1 == f[i][j]) {
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ans += digits[i - 1] + '0';
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j = k;
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}
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}
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int i = 0;
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while (i < ans.size() - 1 && ans[i] == '0') {
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++i;
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}
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return ans.substr(i);
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}
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};
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```
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### **Go**
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```go
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func largestMultipleOfThree(digits []int) string {
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sort.Ints(digits)
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n := len(digits)
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const inf = 1 << 30
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f := make([][]int, n+1)
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for i := range f {
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f[i] = make([]int, 3)
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for j := range f[i] {
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f[i][j] = -inf
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}
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}
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f[0][0] = 0
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for i := 1; i <= n; i++ {
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for j := 0; j < 3; j++ {
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f[i][j] = max(f[i-1][j], f[i-1][(j-digits[i-1]%3+3)%3]+1)
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}
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}
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if f[n][0] <= 0 {
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return ""
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}
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ans := []byte{}
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for i, j := n, 0; i > 0; i-- {
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k := (j - digits[i-1]%3 + 3) % 3
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if f[i][j] == f[i-1][k]+1 {
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ans = append(ans, byte('0'+digits[i-1]))
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j = k
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}
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}
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i := 0
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for i < len(ans)-1 && ans[i] == '0' {
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i++
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}
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return string(ans[i:])
215+
}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **TypeScript**
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```ts
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function largestMultipleOfThree(digits: number[]): string {
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digits.sort((a, b) => a - b);
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const n = digits.length;
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const f: number[][] = new Array(n + 1).fill(0).map(() => new Array(3).fill(-Infinity));
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f[0][0] = 0;
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for (let i = 1; i <= n; ++i) {
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for (let j = 0; j < 3; ++j) {
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f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - (digits[i - 1] % 3) + 3) % 3] + 1);
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}
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}
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if (f[n][0] <= 0) {
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return '';
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}
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const arr: number[] = [];
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for (let i = n, j = 0; i; --i) {
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const k = (j - (digits[i - 1] % 3) + 3) % 3;
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if (f[i - 1][k] + 1 === f[i][j]) {
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arr.push(digits[i - 1]);
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j = k;
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}
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}
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let i = 0;
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while (i < arr.length - 1 && arr[i] === 0) {
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++i;
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}
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return arr.slice(i).join('');
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}
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```
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### **...**

‎solution/1300-1399/1363.Largest Multiple of Three/README_EN.md‎

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### **Python3**
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```python
48-
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class Solution:
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def largestMultipleOfThree(self, digits: List[int]) -> str:
50+
digits.sort()
51+
n = len(digits)
52+
f = [[-inf] * 3 for _ in range(n + 1)]
53+
f[0][0] = 0
54+
for i, x in enumerate(digits, 1):
55+
for j in range(3):
56+
f[i][j] = max(f[i - 1][j], f[i - 1][(j - x % 3 + 3) % 3] + 1)
57+
if f[n][0] <= 0:
58+
return ""
59+
arr = []
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j = 0
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for i in range(n, 0, -1):
62+
k = (j - digits[i - 1] % 3 + 3) % 3
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if f[i - 1][k] + 1 == f[i][j]:
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arr.append(digits[i - 1])
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j = k
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i = 0
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while i < len(arr) - 1 and arr[i] == 0:
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i += 1
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return "".join(map(str, arr[i:]))
4970
```
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### **Java**
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```java
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class Solution {
76+
public String largestMultipleOfThree(int[] digits) {
77+
Arrays.sort(digits);
78+
int n = digits.length;
79+
int[][] f = new int[n + 1][3];
80+
final int inf = 1 << 30;
81+
for (var g : f) {
82+
Arrays.fill(g, -inf);
83+
}
84+
f[0][0] = 0;
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for (int i = 1; i <= n; ++i) {
86+
for (int j = 0; j < 3; ++j) {
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f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - digits[i - 1] % 3 + 3) % 3] + 1);
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}
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}
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if (f[n][0] <= 0) {
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return "";
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}
93+
StringBuilder sb = new StringBuilder();
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for (int i = n, j = 0; i > 0; --i) {
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int k = (j - digits[i - 1] % 3 + 3) % 3;
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if (f[i - 1][k] + 1 == f[i][j]) {
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sb.append(digits[i - 1]);
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j = k;
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}
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}
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int i = 0;
102+
while (i < sb.length() - 1 && sb.charAt(i) == '0') {
103+
++i;
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}
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return sb.substring(i);
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}
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}
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```
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### **C++**
111+
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```cpp
113+
class Solution {
114+
public:
115+
string largestMultipleOfThree(vector<int>& digits) {
116+
sort(digits.begin(), digits.end());
117+
int n = digits.size();
118+
int f[n + 1][3];
119+
memset(f, -0x3f, sizeof(f));
120+
f[0][0] = 0;
121+
for (int i = 1; i <= n; ++i) {
122+
for (int j = 0; j < 3; ++j) {
123+
f[i][j] = max(f[i - 1][j], f[i - 1][(j - digits[i - 1] % 3 + 3) % 3] + 1);
124+
}
125+
}
126+
if (f[n][0] <= 0) {
127+
return "";
128+
}
129+
string ans;
130+
for (int i = n, j = 0; i; --i) {
131+
int k = (j - digits[i - 1] % 3 + 3) % 3;
132+
if (f[i - 1][k] + 1 == f[i][j]) {
133+
ans += digits[i - 1] + '0';
134+
j = k;
135+
}
136+
}
137+
int i = 0;
138+
while (i < ans.size() - 1 && ans[i] == '0') {
139+
++i;
140+
}
141+
return ans.substr(i);
142+
}
143+
};
144+
```
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### **Go**
147+
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```go
149+
func largestMultipleOfThree(digits []int) string {
150+
sort.Ints(digits)
151+
n := len(digits)
152+
const inf = 1 << 30
153+
f := make([][]int, n+1)
154+
for i := range f {
155+
f[i] = make([]int, 3)
156+
for j := range f[i] {
157+
f[i][j] = -inf
158+
}
159+
}
160+
f[0][0] = 0
161+
for i := 1; i <= n; i++ {
162+
for j := 0; j < 3; j++ {
163+
f[i][j] = max(f[i-1][j], f[i-1][(j-digits[i-1]%3+3)%3]+1)
164+
}
165+
}
166+
if f[n][0] <= 0 {
167+
return ""
168+
}
169+
ans := []byte{}
170+
for i, j := n, 0; i > 0; i-- {
171+
k := (j - digits[i-1]%3 + 3) % 3
172+
if f[i][j] == f[i-1][k]+1 {
173+
ans = append(ans, byte('0'+digits[i-1]))
174+
j = k
175+
}
176+
}
177+
i := 0
178+
for i < len(ans)-1 && ans[i] == '0' {
179+
i++
180+
}
181+
return string(ans[i:])
182+
}
183+
184+
func max(a, b int) int {
185+
if a > b {
186+
return a
187+
}
188+
return b
189+
}
190+
```
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192+
### **TypeScript**
193+
194+
```ts
195+
function largestMultipleOfThree(digits: number[]): string {
196+
digits.sort((a, b) => a - b);
197+
const n = digits.length;
198+
const f: number[][] = new Array(n + 1).fill(0).map(() => new Array(3).fill(-Infinity));
199+
f[0][0] = 0;
200+
for (let i = 1; i <= n; ++i) {
201+
for (let j = 0; j < 3; ++j) {
202+
f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - (digits[i - 1] % 3) + 3) % 3] + 1);
203+
}
204+
}
205+
if (f[n][0] <= 0) {
206+
return '';
207+
}
208+
const arr: number[] = [];
209+
for (let i = n, j = 0; i; --i) {
210+
const k = (j - (digits[i - 1] % 3) + 3) % 3;
211+
if (f[i - 1][k] + 1 === f[i][j]) {
212+
arr.push(digits[i - 1]);
213+
j = k;
214+
}
215+
}
216+
let i = 0;
217+
while (i < arr.length - 1 && arr[i] === 0) {
218+
++i;
219+
}
220+
return arr.slice(i).join('');
221+
}
55222
```
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### **...**

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