不会死的二分模板
int findPosition(vector<int> &nums,int target)
{
int left=0;
int right=nums.size()-1;
//结束条件为left和right相邻
while(left+1<right)
{
int mid=left+(right-left)/2;
if(a[mid]==target)
right=mid;
else if(a[mid]<target)
left=mid;
else if(a[mid>target])
right=mid;
}
if(a[left]==target)
return left;
if(a[right]==target)
return right;
return -1;
}
int findPosition(vector<int> &nums,int target)
{
int left=0;
int right=nums.size()-1;
//结束条件为left和right相邻
while(left+1<right)
{
int mid=left+(right-left)/2;
if(a[mid]==target)
left=mid;
else if(a[mid]<target)
left=mid;
else if(a[mid>target])
right=mid;
}
if(a[right]==target)
return right;
if(a[left]==target)
return left;
return -1;
}
题目
题目一:457 · 经典二分查找问题 - LintCode
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class Solution {
public:
int findPosition(vector<int> &nums, int target) {
// write your code here
if(nums.size()==0)
return -1;
int left=0;
int right=nums.size()-1;
while(left+1<right)
{
int mid=left+(right-left)/2;
if(nums[mid]==target)
{
right=mid;
}
else if(nums[mid]<target)
{
left=mid;
}
else if(nums[mid]>target)
{
right=mid;
}
}
if(nums[left]==target)
return left;
if(nums[right]==target)
return right;
return -1;
}
};
题目二:458 · 目标最后位置 - LintCode)
class Solution {
public:
/**
* @param nums: An integer array sorted in ascending order
* @param target: An integer
* @return: An integer
*/
int lastPosition(vector<int> &nums, int target) {
// write your code here
int left=0;
int right=nums.size()-1;
if(nums.size()==0)
return -1;
while(left+1<right)
{
int mid=left+(right-left)/2;
if(nums[mid]==target)
left=mid;
else if(nums[mid]<target)
left=mid;
else if(nums[mid]>target)
right=mid;
}
if(nums[right]==target)
return right;
if(nums[left]==target)
return left;
return -1;
}
};
题目三：65 · 两个排序数组的中位数 - LintCode
class Solution {
public:
/**
* @param a: An integer array
* @param b: An integer array
* @return: a double whose format is *.5 or *.0
*/
double found(vector<int> &a, vector<int> &b,int starta,int startb,int k)
{
//如果A数组里面没有数
if(starta>=a.size())
return b[startb+k-1];
//如果B数组里面没有数
if(startb>=b.size())
return a[starta+k-1];
//递归的退出条件,如果找最小的数就找两个数组第一个数的较小数
if(k==1)
return min(a[starta],b[startb]);
//如果一个数组太短，那么删除的一定是长的数组的前k/2
//所以如果数组长度不够，比较值设置为INT_MAX，删除的就是另一个数组的数
int keya=starta+k/2-1>=a.size()?INT_MAX:a[starta+k/2-1];
int keyb=startb+k/2-1>=b.size()?INT_MAX:b[startb+k/2-1];
if(keya>keyb)
return found(a,b,starta,startb+k/2,k-k/2);
else
return found(a,b,starta+k/2,startb,k-k/2);
}
double findMedianSortedArrays(vector<int> &a, vector<int> &b) {
// write your code here
int sum=a.size()+b.size();
double ret;
if(sum&1)
ret=found(a,b,0,0,sum/2+1);
else
ret=(found(a,b,0,0,sum/2)+found(a,b,0,0,sum/2+1))/2;
return ret;
}
};
题目四：28 · 搜索二维矩阵 - LintCode
class Solution {
public:
/**
* @param matrix: matrix, a list of lists of integers
* @param target: An integer
* @return: a boolean, indicate whether matrix contains target
*/
bool searchMatrix(vector<vector<int>> &matrix, int target) {
// write your code here
int n=matrix.size();
if(maxtrix==NULL||n==0) return false;
int m=matrix[0].size();
if(maxtrix[0]==NULL||m==0) return false;
//进行二分查找
int left=0;
int right=n*m-1;
while(left+1<right)
{
int mid=left+(right-left)/2;
int row=mid/m;
int col=mid%m;
if(matrix[row][col]==target)
return true;
else if(matrix[row][col]>target)
right=mid;
else if(matrix[row][col]<target)
left=mid;
}
if(matrix[left/m][left%m]==target)
return true;
if(matrix[right/m][right%m]==target)
return true;
return false;
}
};
题目四：60 · 搜索插入位置 - LintCode
class Solution {
public:
/**
* @param a: an integer sorted array
* @param target: an integer to be inserted
* @return: An integer
*/
int searchInsert(vector<int> &a, int target) {
// write your code here
if(a.size()==0)
return 0;
int left=0;
int right=a.size()-1;
while(left+1<right)
{
int mid=left+(right-left)/2;
if(a[mid]==target)
return mid;
else if(a[mid]>target)
right=mid;
else if(a[mid]<target)
left=mid;
}
if(target>a[right])
return right+1;
if(target>a[left])
return right;
return 0;
}
};
题目五：447 · 在大数组中查找 - LintCode
class Solution {
public:
int searchBigSortedArray(ArrayReader * reader, int target) {
int left=0,right=1;
while(reader->get(right)<target)
right<<=1;
//在区间内二分查找
while(left+1<right)
{
int mid=left+(right-left)/2;
if(reader->get(mid)<target)
left=mid;
else
right=mid;
}
if(reader->get(left)==target)
return left;
if(reader->get(right)==target)
return right;
return -1;
}
};
题目六：159 · 寻找旋转排序数组中的最小值 - LintCode
class Solution {
public:
int findMin(vector<int> &nums) {
int left=0;
int right=nums.size()-1;
while(left+1<right)
{
int mid=left+(right-left)/2;
if(nums[mid]>nums[right])
left=mid;
else
right=mid;
}
//相邻时退出
return min(nums[left],nums[right]);
}
};
题目七：62 · 搜索旋转排序数组 - LintCode
class Solution {
public:
int search(vector<int> &a, int target) {
int left=0,right=a.size()-1;
if(left>right)
return -1;
while(left+1<right)
{
int mid=left+(right-left)/2;
//如果mid落在了左区间
if(a[mid]>a[left])
{
//如果target在mid的左边并且不再右区间
if(target<a[mid]&&a[left]<=target)
right=mid;
else
left=mid;
}
else
{
if(target>a[mid]&&target<=a[right])
left=mid;
else
right=mid;
}
}
if(a[left]==target)
return left;
if(a[right]==target)
return right;
return -1;
}
};
题目八：183 · 木材加工 - LintCode
class Solution {
public:
//判断是否长度是否合适
bool cut(vector<int>&l,int len,int k)
{
int count=0;
for(int i=0;i<l.size();i++)
{
count+=l[i]/len;
}
if(count>=k)
return true;
else
return false;
}
int woodCut(vector<int> &l, int k) {
if(l.size()==0)
return 0;
int maxl=0;
for(int i=0;i<l.size();i++)
{
maxl=max(maxl,l[i]);
}
int start=0,end=maxl;
while(start+1<end)
{
int mid=start+(end-start)/2;
if(cut(l,mid,k))
{
start=mid;
}
else
end=mid;
}
if(cut(l,end,k)) return end;
return start;
}
};
