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Commit ff87889

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feat: add solutions to lc problem: No.1089
No.1089.Duplicate Zeros
1 parent 24613ee commit ff87889

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4 files changed

+182
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lines changed

4 files changed

+182
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lines changed

‎solution/1000-1099/1089.Duplicate Zeros/README.md‎

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@@ -41,7 +41,21 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:双指针**
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由于是原地修改,所以不能直接将 0 的后一位直接修改为 0,这会丢失元素数据。
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若选择插入,则会导致元素位置调整,时间复杂度偏高。
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**方法一:模拟**
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开辟一个等长数组,将 `arr` 复刻一份,再进行简单模拟即可。
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- 时间复杂度:$O(n)$。
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- 空间复杂度:$O(n)$。
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**方法二:双指针**
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- 时间复杂度:$O(n)$。
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- 空间复杂度:$O(1)$。
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<!-- tabs:start -->
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@@ -161,6 +175,65 @@ func duplicateZeros(arr []int) {
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}
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```
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### **C**
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```c
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void duplicateZeros(int* arr, int arrSize){
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int i = 0;
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int j = 0;
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while (j < arrSize) {
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if (arr[i] == 0) {
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j++;
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}
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i++;
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j++;
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}
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i--;
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j--;
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while (i >= 0) {
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if (arr[i] == 0) {
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if (j < arrSize) {
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arr[j] = arr[i];
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}
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j--;
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}
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arr[j] = arr[i];
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i--;
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j--;
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}
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn duplicate_zeros(arr: &mut Vec<i32>) {
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let n = arr.len();
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let mut i = 0;
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let mut j = 0;
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while j < n {
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if arr[i] == 0 {
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j += 1;
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}
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j += 1;
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i += 1;
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}
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while i > 0 {
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if arr[i - 1] == 0 {
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if j <= n {
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arr[j - 1] = arr[i - 1];
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}
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j -= 1;
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}
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arr[j - 1] = arr[i - 1];
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i -= 1;
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j -= 1;
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}
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}
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}
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```
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### **...**
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```

‎solution/1000-1099/1089.Duplicate Zeros/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -149,6 +149,65 @@ func duplicateZeros(arr []int) {
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}
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```
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### **C**
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```c
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void duplicateZeros(int* arr, int arrSize){
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int i = 0;
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int j = 0;
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while (j < arrSize) {
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if (arr[i] == 0) {
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j++;
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}
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i++;
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j++;
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}
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i--;
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j--;
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while (i >= 0) {
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if (arr[i] == 0) {
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if (j < arrSize) {
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arr[j] = arr[i];
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}
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j--;
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}
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arr[j] = arr[i];
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i--;
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j--;
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}
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}
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```
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### **Rust**
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```rust
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impl Solution {
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pub fn duplicate_zeros(arr: &mut Vec<i32>) {
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let n = arr.len();
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let mut i = 0;
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let mut j = 0;
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while j < n {
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if arr[i] == 0 {
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j += 1;
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}
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j += 1;
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i += 1;
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}
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while i > 0 {
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if arr[i - 1] == 0 {
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if j <= n {
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arr[j - 1] = arr[i - 1];
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}
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j -= 1;
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}
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arr[j - 1] = arr[i - 1];
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i -= 1;
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j -= 1;
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}
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}
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}
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```
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### **...**
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```
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
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void duplicateZeros(int* arr, int arrSize){
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int i = 0;
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int j = 0;
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while (j < arrSize) {
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if (arr[i] == 0) {
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j++;
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}
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i++;
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j++;
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}
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i--;
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j--;
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while (i >= 0) {
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if (arr[i] == 0) {
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if (j < arrSize) {
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arr[j] = arr[i];
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}
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j--;
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}
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arr[j] = arr[i];
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i--;
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j--;
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}
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}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,25 @@
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impl Solution {
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pub fn duplicate_zeros(arr: &mut Vec<i32>) {
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let n = arr.len();
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let mut i = 0;
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let mut j = 0;
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while j < n {
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if arr[i] == 0 {
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j += 1;
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}
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j += 1;
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i += 1;
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}
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while i > 0 {
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if arr[i - 1] == 0 {
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if j <= n {
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arr[j - 1] = arr[i - 1];
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}
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j -= 1;
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}
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arr[j - 1] = arr[i - 1];
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i -= 1;
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j -= 1;
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}
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}
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}

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