---
comments: true
difficulty: 中等
edit_url: https://github.com/doocs/leetcode/edit/main/solution/0000-0099/0034.Find%20First%20and%20Last%20Position%20of%20Element%20in%20Sorted%20Array/README.md
tags:
    - 数组
    - 二分查找
---

<!-- problem:start -->

# [34. 在排序数组中查找元素的第一个和最后一个位置](https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array)

[English Version](/solution/0000-0099/0034.Find%20First%20and%20Last%20Position%20of%20Element%20in%20Sorted%20Array/README_EN.md)

## 题目描述

<!-- description:start -->

<p>给你一个按照非递减顺序排列的整数数组 <code>nums</code>，和一个目标值 <code>target</code>。请你找出给定目标值在数组中的开始位置和结束位置。</p>

<p>如果数组中不存在目标值 <code>target</code>，返回&nbsp;<code>[-1, -1]</code>。</p>

<p>你必须设计并实现时间复杂度为&nbsp;<code>O(log n)</code>&nbsp;的算法解决此问题。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<pre>
<strong>输入：</strong>nums = [<code>5,7,7,8,8,10]</code>, target = 8
<strong>输出：</strong>[3,4]</pre>

<p><strong>示例&nbsp;2：</strong></p>

<pre>
<strong>输入：</strong>nums = [<code>5,7,7,8,8,10]</code>, target = 6
<strong>输出：</strong>[-1,-1]</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>nums = [], target = 0
<strong>输出：</strong>[-1,-1]</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>0 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup>&nbsp;&lt;= nums[i]&nbsp;&lt;= 10<sup>9</sup></code></li>
	<li><code>nums</code>&nbsp;是一个非递减数组</li>
	<li><code>-10<sup>9</sup>&nbsp;&lt;= target&nbsp;&lt;= 10<sup>9</sup></code></li>
</ul>

<!-- description:end -->

## 解法

<!-- solution:start -->

### 方法一：二分查找

我们可以进行两次二分查找，分别查找出左边界和右边界。

时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。其中 $n$ 是数组 $nums$ 的长度。

以下是二分查找的两个通用模板：

模板 1：

```java
boolean check(int x) {
}

int search(int left, int right) {
    while (left < right) {
        int mid = (left + right) >> 1;
        if (check(mid)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}
```

模板 2：

```java
boolean check(int x) {
}

int search(int left, int right) {
    while (left < right) {
        int mid = (left + right + 1) >> 1;
        if (check(mid)) {
            left = mid;
        } else {
            right = mid - 1;
        }
    }
    return left;
}
```

做二分题目时，可以按照以下套路：

1. 写出循环条件 $left < right$；
1. 循环体内，不妨先写 $mid = \lfloor \frac{left + right}{2} \rfloor$；
1. 根据具体题目，实现 $check()$ 函数（有时很简单的逻辑，可以不定义 $check$），想一下究竟要用 $right = mid$（模板 $1$） 还是 $left = mid$（模板 $2$）；
       - 如果 $right = mid$，那么写出 else 语句 $left = mid + 1$，并且不需要更改 mid 的计算，即保持 $mid = \lfloor \frac{left + right}{2} \rfloor$；
       - 如果 $left = mid$，那么写出 else 语句 $right = mid - 1$，并且在 $mid$ 计算时补充 +1，即 $mid = \lfloor \frac{left + right + 1}{2} \rfloor$；
1. 循环结束时， $left$ 与 $right$ 相等。

注意，这两个模板的优点是始终保持答案位于二分区间内，二分结束条件对应的值恰好在答案所处的位置。 对于可能无解的情况，只要判断二分结束后的 $left$ 或者 $right$ 是否满足题意即可。

<!-- tabs:start -->

#### Python3

```python
class Solution:
    def searchRange(self, nums: List[int], target: int) -> List[int]:
        l = bisect_left(nums, target)
        r = bisect_left(nums, target + 1)
        return [-1, -1] if l == r else [l, r - 1]
```

#### Java

```java
class Solution {
    public int[] searchRange(int[] nums, int target) {
        int l = search(nums, target);
        int r = search(nums, target + 1);
        return l == r ? new int[] {-1, -1} : new int[] {l, r - 1};
    }

    private int search(int[] nums, int x) {
        int left = 0, right = nums.length;
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (nums[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}
```

#### C++

```cpp
class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int l = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
        int r = lower_bound(nums.begin(), nums.end(), target + 1) - nums.begin();
        if (l == r) return {-1, -1};
        return {l, r - 1};
    }
};
```

#### Go

```go
func searchRange(nums []int, target int) []int {
	l := sort.SearchInts(nums, target)
	r := sort.SearchInts(nums, target+1)
	if l == r {
		return []int{-1, -1}
	}
	return []int{l, r - 1}
}
```

#### TypeScript

```ts
function searchRange(nums: number[], target: number): number[] {
    const search = (x: number): number => {
        let [left, right] = [0, nums.length];
        while (left < right) {
            const mid = (left + right) >> 1;
            if (nums[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    };
    const l = search(target);
    const r = search(target + 1);
    return l === r ? [-1, -1] : [l, r - 1];
}
```

#### Rust

```rust
impl Solution {
    pub fn search_range(nums: Vec<i32>, target: i32) -> Vec<i32> {
        let n = nums.len();
        let search = |x| {
            let mut left = 0;
            let mut right = n;
            while left < right {
                let mid = left + (right - left) / 2;
                if nums[mid] < x {
                    left = mid + 1;
                } else {
                    right = mid;
                }
            }
            left
        };
        let l = search(target);
        let r = search(target + 1);
        if l == r {
            return vec![-1, -1];
        }
        vec![l as i32, (r - 1) as i32]
    }
}
```

#### JavaScript

```js
/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var searchRange = function (nums, target) {
    function search(x) {
        let left = 0,
            right = nums.length;
        while (left < right) {
            const mid = (left + right) >> 1;
            if (nums[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
    const l = search(target);
    const r = search(target + 1);
    return l == r ? [-1, -1] : [l, r - 1];
};
```

#### PHP

```php
class Solution {
    /**
     * @param integer[] $nums
     * @param integer $target
     * @return integer[]
     */

    function searchRange($nums, $target) {
        $min = -1;
        $max = -1;
        foreach ($nums as $key => $value) {
            if ($value == $target) {
                if ($min == -1) {
                    $min = $key;
                }

                if ($key > $max) {
                    $max = $key;
                }
            }
        }
        return [$min, $max];
    }
}
```

#### Kotlin

```kotlin
class Solution {
    fun searchRange(nums: IntArray, target: Int): IntArray {
        val left = this.search(nums, target)
        val right = this.search(nums, target + 1)
        return if (left == right) intArrayOf(-1, -1) else intArrayOf(left, right - 1)
    }

    private fun search(nums: IntArray, target: Int): Int {
        var left = 0
        var right = nums.size
        while (left < right) {
            val middle = (left + right) / 2
            if (nums[middle] < target) {
                left = middle + 1
            } else {
                right = middle
            }
        }
        return left
    }
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->