1. 思想
尋找一個標準,可以使原始數據具有兩段性
2. 模板
情況一:
按照某種性質將數據分爲兩類,target在後半部分
- 若此時mid > target,則R = mid;否則L = mid + 1
int bsearch_1(int l, int r)
{
while(l < r)
{
int mid = l + r >> 1;
if(check(mid)) r = mid;
else l = mid + 1;
}
return l;
}
情況二:
按照某種性質將數據分爲兩類,target在前半部分
- 若此時mid > target,則R = mid - 1;否則 L = mid
int bsearch_2(int l, int r)
{
while (l < r)
{
int mid = l + r + 1 >> 1; //注意
if (check(mid)) l = mid;
else r = mid - 1;
}
return l;
}
3. 題目
//模板2
class Solution {
public:
int mySqrt(int x) {
int l = 0;
int r = x;
while(l < r)
{
int mid = (long long)l+r+1>>1;
if((long long)mid*mid > x)
r = mid - 1;
else
l = mid;
}
return l;
}
};
//模板一
class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
int l = 0;
int r = nums.size();
while(l < r)
{
int mid = l+r>>1;
if(nums[mid] >= target)
r = mid;
else
l = mid + 1;
}
return l;
}
};
特殊題
class Solution {
public:
int findPeakElement(vector<int>& nums) {
int l = 0;
int r = nums.size()-1;
while(l < r)
{
int mid = l+r>>1;
if(nums[mid] > nums[mid+1]) //當前值 比 下一個值大,左邊肯定有峯值
r = mid;
else
l = mid + 1;
}
return l;
}
};
將每個數的取值的區間[1, n]劃分成[1, n/2]和[n/2+1, n]兩個子區間,然後分別統計兩個區間中數的個數。若在區間[1, n/2]中數的個數大於區間長度,則此區間一定有重複的數。
class Solution {
public:
int findDuplicate(vector<int>& nums) {
int n = nums.size() - 1;
int l = 1, r = n;
while (l < r)
{
int mid = l + r >> 1;
int cnt = 0;
for (int x : nums)
if(l <= x && x <= mid)
cnt++;
if (cnt > mid - l + 1) r = mid;
else l = mid + 1;
}
return r;
}
};