就是为了干倒二分题目。
题目
475. Heaters
Winter is coming! Your first job during the contest is to design a standard heater with fixed warm radius to warm all the houses.
Now, you are given positions of houses and heaters on a horizontal line, find out minimum radius of heaters so that all houses could be covered by those heaters.
So, your input will be the positions of houses and heaters seperately, and your expected output will be the minimum radius standard of heaters.
Note:
Numbers of houses and heaters you are given are non-negative and will not exceed 25000.
Positions of houses and heaters you are given are non-negative and will not exceed 10^9.
As long as a house is in the heaters' warm radius range, it can be warmed.
All the heaters follow your radius standard and the warm radius will the same.
Example 1:
Input: [1,2,3],[2]
Output: 1
Explanation: The only heater was placed in the position 2, and if we use the radius 1 standard, then all the houses can be warmed.
Example 2:
Input: [1,2,3,4],[1,4]
Output: 1
Explanation: The two heater was placed in the position 1 and 4. We need to use radius 1 standard, then all the houses can be warmed.
- 将房屋和暖气的位置分别从小到大排序。
- 假设得到了加热半径 r，即可在O(n+m)的时间内，判断是否所有的房屋都得到了暖气，具体为，逐一枚举房屋，然后判断是否有暖气覆盖，由于房屋和暖气的坐标都是单调的，所以第i+1个房屋不会使用坐标小于第i个房屋的暖气位置。
- 加热半径也是单调的，故可以使用二分来加速寻找最小的半径。
import java.util.*;
class Main {
//定义域：半径长度
//值域：给房所有房子供暖
//最优解：可以给房子供暖的半径中求一条最短的。low_bound
public int findRadius(int[] houses, int[] heaters) {
Arrays.sort(heaters);
Arrays.sort(houses);
if (houses.length==0)
return 0;
int l=0;
int r=Math.max(houses[houses.length-1],heaters[heaters.length-1]);
while (l < r)
{
int mid = l + r >> 1;
if (check(mid,houses,heaters)) r = mid;//半径长度太长
else l = mid + 1;//半径长度太短
}
return l;
}
boolean check(int r,int[] houses, int[] heaters){
int n = heaters.length;
for (int i = 0, j = 0; i < houses.length; i++) {
while (j < n && Math.abs(houses[i] - heaters[j]) > r)
j++;
if (j == n)
return false;
}
return true;
}
public static void main(String[] args) {
}
}
题目
69. Sqrt(x)
Implement int sqrt(int x).
Compute and return the square root of x, where x is guaranteed to be a non-negative integer.
Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.
Example 1:
Input: 4
Output: 2
Example 2:
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since
the decimal part is truncated, 2 is returned.
定义域：输入x
值域：x的开方
最优解：mid*mid<=x去mid最大的那个
class Solution {
public int mySqrt(int x) {
return bsearch_2(0, x, x);
}
int bsearch_2(int l, int r, int x) {
while (l < r) {
int mid = l + r + 1 >> 1;
if (mid <= x / mid) l = mid;
else r = mid - 1;
}
return l;
}
}
题目
875. Koko Eating Bananas
Koko loves to eat bananas. There are N piles of bananas, the i-th pile has piles[i] bananas. The guards have gone and will come back in H hours.
Koko can decide her bananas-per-hour eating speed of K. Each hour, she chooses some pile of bananas, and eats K bananas from that pile. If the pile has less than K bananas, she eats all of them instead, and won't eat any more bananas during this hour.
Koko likes to eat slowly, but still wants to finish eating all the bananas before the guards come back.
Return the minimum integer K such that she can eat all the bananas within H hours.
Example 1:
Input: piles = [3,6,7,11], H = 8
Output: 4
Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30
Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23
//定义域：速度
//值域：是否可以吃掉所有的香蕉
//最优解：在所以吃掉香蕉的速度中取最小 low_bound
public int minEatingSpeed(int[] piles, int H) {
int maxspeed = 0;
int a = 0;
for (int i = 0; i < piles.length; i++) {
a = Math.max(a, piles[i]);
}
maxspeed = a * piles.length / H;
return bsearch_1(1, 1_000_000_000, piles, H);
}
int bsearch_1(int l, int r, int[] piles, int H) {
while (l < r) {
int mid = l + r >> 1;//同l+( r-l )/2;
if (check(mid, piles, H)) r = mid;
else l = mid + 1;
}
return l;
}
private boolean check(int mid, int[] piles, int H) {
int sum = 0;
for (int i = 0; i < piles.length; i++) {
if (piles[i] % mid == 0)
sum += piles[i] / mid;
else
sum += piles[i] / mid + 1;
// sum+=(piles[i]-1)/mid+1;
}
if (sum > H)//速度太慢了
return false;
return true;//速度ok包括速度刚好应该在r这边
}