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Commit 808ddd9

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feat: add solutions to lc problem: No.704 (doocs#2518)
No.0704.Binary Search
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‎solution/0700-0799/0704.Binary Search/README.md‎

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## 解法
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### 方法一
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### 方法一:二分查找
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我们定义二分查找的左边界 $left=0,ドル右边界 $right=n-1$。
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每一次循环,我们计算中间位置 $mid=(left+right)/2,ドル然后判断 $nums[mid]$ 和 $target$ 的大小关系:
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- 如果 $nums[mid] \geq target,ドル则说明 $target$ 在 $[left, mid]$ 之间,我们将 $right$ 更新为 $mid$;
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- 否则,说明 $target$ 在 $[mid+1, right]$ 之间,我们将 $left$ 更新为 $mid+1$。
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当 $left \geq right$ 时,我们判断 $nums[left]$ 是否等于 $target,ドル如果等于则返回 $left,ドル否则返回 $-1$。
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时间复杂度 $O(\log n),ドル其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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};
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```
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<!-- tabs:end -->
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### 方法二
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<!-- tabs:start -->
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```rust
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use std::cmp::Ordering;
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impl Solution {
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fn binary_search(nums: Vec<i32>, target: i32, l: usize, r: usize) -> i32 {
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if l == r {
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return if nums[l] == target { l as i32 } else { -1 };
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}
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let mid = (l + r) >> 1;
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match nums[mid].cmp(&target) {
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Ordering::Less => Self::binary_search(nums, target, mid + 1, r),
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Ordering::Greater => Self::binary_search(nums, target, l, mid),
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Ordering::Equal => mid as i32,
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```cs
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public class Solution {
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public int Search(int[] nums, int target) {
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int left = 0, right = nums.Length - 1;
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while (left < right) {
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int mid = (left + right) >> 1;
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if (nums[mid] >= target) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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}
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pub fn search(nums: Vec<i32>, target: i32) -> i32 {
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let r = nums.len() - 1;
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Self::binary_search(nums, target, 0, r)
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return nums[left] == target ? left : -1;
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}
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}
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```

‎solution/0700-0799/0704.Binary Search/README_EN.md‎

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## Solutions
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### Solution 1
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### Solution 1: Binary Search
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We define the left boundary of the binary search as $left=0,ドル and the right boundary as $right=n-1$.
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In each iteration, we calculate the middle position $mid=(left+right)/2,ドル and then compare the size of $nums[mid]$ and $target$:
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- If $nums[mid] \geq target,ドル it means that $target$ is in the interval $[left, mid],ドル so we update $right$ to $mid$;
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- Otherwise, $target$ is in the interval $[mid+1, right],ドル so we update $left$ to $mid+1$.
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When $left \geq right,ドル we check if $nums[left]$ equals $target$. If it does, we return $left,ドル otherwise, we return $-1$.
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The time complexity is $O(\log n),ドル where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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};
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```
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<!-- tabs:end -->
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### Solution 2
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<!-- tabs:start -->
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```rust
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use std::cmp::Ordering;
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impl Solution {
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fn binary_search(nums: Vec<i32>, target: i32, l: usize, r: usize) -> i32 {
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if l == r {
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return if nums[l] == target { l as i32 } else { -1 };
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}
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let mid = (l + r) >> 1;
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match nums[mid].cmp(&target) {
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Ordering::Less => Self::binary_search(nums, target, mid + 1, r),
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Ordering::Greater => Self::binary_search(nums, target, l, mid),
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Ordering::Equal => mid as i32,
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```cs
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public class Solution {
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public int Search(int[] nums, int target) {
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int left = 0, right = nums.Length - 1;
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while (left < right) {
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int mid = (left + right) >> 1;
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if (nums[mid] >= target) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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}
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pub fn search(nums: Vec<i32>, target: i32) -> i32 {
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let r = nums.len() - 1;
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Self::binary_search(nums, target, 0, r)
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return nums[left] == target ? left : -1;
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}
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}
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```
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public class Solution {
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public int Search(int[] nums, int target) {
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int left = 0, right = nums.Length - 1;
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while (left < right) {
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int mid = (left + right) >> 1;
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if (nums[mid] >= target) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return nums[left] == target ? left : -1;
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}
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}

‎solution/0700-0799/0704.Binary Search/Solution2.rs‎

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