二分查找（小于等于指定数的最小值）：求解《875. 爱吃香蕉的珂珂》
2022-06-07 14:12:36
二分查找（小于等于指定数的最小值），求解《875. 爱吃香蕉的珂珂》
例题
875. 爱吃香蕉的珂珂
珂珂喜欢吃香蕉。这里有 n 堆香蕉，第 i 堆中有 piles[i] 根香蕉。警卫已经离开了，将在 h 小时后回来。
珂珂可以决定她吃香蕉的速度 k （单位：根/小时）。每个小时，她将会选择一堆香蕉，从中吃掉 k 根。如果这堆香蕉少于 k 根，她将吃掉这堆的所有香蕉，然后这一小时内不会再吃更多的香蕉。
珂珂喜欢慢慢吃，但仍然想在警卫回来前吃掉所有的香蕉。
返回她可以在 h 小时内吃掉所有香蕉的最小速度 k（k 为整数）。
示例 1：
输入：piles = [3,6,7,11], h = 8
输出：4
答案
二分查找
var minEatingSpeed = function(piles, h) {
const n = piles.length
let max = 1, sum = 0
for (let i = 0; i < n; i++) {
sum += piles[i]
if (piles[i] > max) max = piles[i]
}
let l = sum / h | 0, r = max
while (l <= r) {
const m = l + r >>> 1
let time = 0
for (let i = 0; i < n; i++) {
time += Math.ceil(piles[i] / m)
}
if (time <= h) r = m - 1
else l = m + 1
}
return l
};function minEatingSpeed(piles: number[], h: number): number {
const n = piles.length
let sum = 0, max = 1
for (let i = 0; i < n; i++) {
sum += piles[i]
if (piles[i] > max) max = piles[i]
}
let l = sum / h | 0, r = max
while (l <= r) {
const m = l + r >>> 1
let time = 0
for (let i = 0; i < n; i++) {
time += Math.trunc((piles[i] - 1) / m) + 1
}
if (time <= h) r = m - 1
else l = m + 1
}
return l
};class Solution {
function minEatingSpeed($piles, $h) {
$sum = 0;
$max = 1;
foreach ($piles as $pile) {
$sum += $pile;
if ($pile > $max) $max = $pile;
}
$l = $sum / $h | 0;
$r = $max;
while ($l <= $r) {
$m = $l + $r >> 1;
$time = 0;
if ($m == 0) return 1;
foreach($piles as $pile) {
$time += ceil($pile / $m);
}
if ($time <= $h) $r = $m - 1;
else $l = $m + 1;
}
return $l;
}
}func minEatingSpeed(piles []int, h int) int {
max := 1
for _, pile := range piles {
if pile > max {
max = pile
}
}
return 1 + sort.Search(max, func(m int) bool {
m++
time := 0
for _, pile := range piles {
time += (pile - 1) / m + 1
}
return time <= h
})
}class Solution {
public int minEatingSpeed(int[] piles, int h) {
long sum = 0, max = 1;
for (int pile : piles) {
sum += pile;
if (pile > max) max = pile;
}
long l = sum / h;
long r = max;
while (l <= r) {
long m = l + r >>> 1, time = 0;
if (m == 0) return 1;
for (int pile : piles) {
time += (pile - 1) / m + 1;
}
if (time <= h) r = m - 1;
else l = m + 1;
}
return (int)l;
}
}class Solution:
def minEatingSpeed(self, piles: List[int], h: int) -> int:
l, r = sum(piles) // h, max(piles)
while l <= r:
m, time = l + r >> 1, 0
if m == 0: return 1
for _, pile in enumerate(piles):
time += (pile - 1) // m + 1
if time <= h: r = m - 1
else: l = m + 1
return l