文章目录
二分搜索的泛化问题
首先从问题中抽象出自变量X,就是题目要求取的最值
其次,抽象出映射函数F(X),捋清对应关系,使其单调
最后,计算F(X) == target,返回X
前言
一、力扣875. 爱吃香蕉的珂珂
class Solution {
public int minEatingSpeed(int[] piles, int h) {
int left = 1, right = 1000000000;
while(left <= right){
int mid = left + (right-left)/2;
if(fun(mid,piles) == h){
right = mid - 1;
}else if(fun(mid,piles) > h){
left = mid + 1;
}else{
right = mid - 1;
}
}
return left;
}
public long fun(int k, int[] piles){
long hour = 0;
for(int i = 0; i < piles.length; i ++){
hour += piles[i]/k;
if(piles[i]%k > 0){
hour ++;
}
}
return hour;
}
}
二、力扣1011. 在 D 天内送达包裹的能力
class Solution {
public int shipWithinDays(int[] weights, int days) {
int left = 0, right = 0;
for(int i = 0; i < weights.length; i ++){
left = Math.max(left,weights[i]);
right += weights[i];
}
while(left <= right){
int mid = left + (right - left)/2;
if(fun(mid, weights) == days){
right = mid - 1;
}else if(fun(mid, weights) > days){
left = mid + 1;
}else{
right = mid - 1;
}
}
return left;
}
public int fun(int k, int[] weights){
int day = 0;
for(int i = 0; i < weights.length;){
int cap = k;
while(i < weights.length){
if(cap < weights[i])break;
else cap -= weights[i++];
}
day ++;
}
return day;
}
}
三、力扣410. 分割数组的最大值
class Solution {
public int splitArray(int[] nums, int k) {
int left = 0, right = 0;
for(int i = 0; i < nums.length;i++){
left = Math.max(left,nums[i]);
right += nums[i];
}
while(left <= right){
int mid = left + (right - left)/2;
if(fun(mid, nums) == k){
right = mid - 1;
}else if(fun(mid, nums) > k){
left = mid + 1;
}else{
right = mid - 1;
}
}
return left;
}
public int fun(int x, int[] nums){
int m = 0;
for(int i = 0; i < nums.length;){
int cap = x;
while(i < nums.length){
if(cap < nums[i])break;
else cap -= nums[i];
i ++;
}
m ++;
}
return m;
}
}