给定一个长度为n的非降序排列的整数数组
,和一个目标值
。请找出给定目标值在数组中的开始位置和结束位置。
如果数组中不存在目标值
,返回 [-1, -1]
该题有
的解法
[4,7,7,8,8,10],8
[3,4]
可以在数组中找到整数8,其开始位置为3,结束位置为4
[4,7,7,8,8,10],6
[-1,-1]
不可以在数组中找到整数6,故输出整数-1
struct Solution{
}
use std::cmp::Ordering::{Less, Equal, Greater};
impl Solution {
fn new() -> Self {
Solution{}
}
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param nums int整型一维数组
* @param target int整型
* @return int整型一维数组
*/
pub fn searchRange(&self, nums: Vec<i32>, target: i32) -> Vec<i32> {
// write code here
vec![
Self::floor(&nums, [-1, nums.len() as i32 - 1], target),
Self::ceil(&nums, (-1, nums.len() as i32), target)
]
}
fn floor(v: &Vec<i32>, [l, r]: [i32; 2], t: i32) -> i32 {
// (l, r]
if l >= r {return -1}
let mid = (l + 1) + (r - l - 1) / 2;
// if mid < 0 {}
match t.cmp(&v[mid as usize]) {
Less => Self::floor(v, [l, mid - 1], t),
Greater => Self::floor(v, [mid, r], t),
Equal => {
let r = Self::floor(v, [l, mid - 1], t);
if r == -1 {
mid
} else {r}
}
}
}
fn ceil(v: &Vec<i32>, (l, r): (i32, i32), t: i32) -> i32 {
// (l, r)
if r - l < 2 {
return -1
}
let mid = (l + r)/2;
match v[mid as usize].cmp(&t) {
Greater => Self::ceil(v, (l, mid), t),
Less => Self::ceil(v, (mid, r), t),
Equal => {
let r= Self::ceil(v, (mid, r), t);
if r == -1 {mid} else {r}
}
}
}
}