二分套路
y神模板
b站链接：https://www.bilibili.com/video/BV1Ft41157zW/?spm_id_from=333.1007.top_right_bar_window_history.content.click&vd_source=5500fef761fd3ab451b9bbe518cd5cc6
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模板一
int bsearch_1(int l, int r)
{
while (l < r)
{
int mid = l + r >> 1; //除2 操作
if (check(mid)) r = mid; //check函数即为边界的选择
else l = mid + 1;
}
return l; //l r都可以，跳出while r = l
}
模板二
int bsearch_2(int l, int r)
{
while (l < r)
{
int mid = l + r + 1 >> 1; //注意+1操作
if (check(mid)) l = mid;
else r = mid - 1;
}
return l;
}
实战
lc.x的平方根https://leetcode.cn/problems/sqrtx/description/
class Solution {
public int mySqrt(int x) {
//套y神模板
int l = 0, r = x;
while(l < r){
int mid = (l + r + 1) >> 1;
if(mid <= x / mid){
l = mid;
}else{
r = mid - 1;
}
}
return r;
}
}
lc74 搜索二维矩阵 https://leetcode.cn/problems/search-a-2d-matrix/
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
//使用y神模板
if(matrix == null || matrix.length == 0) return false;
int n = matrix.length, m = matrix[0].length;
int l = 0, r = n * m - 1;
while(l < r){
int mid = (l + r) >> 1;
if(matrix[mid / m][mid % m] >= target){
r = mid;
}else{
l = mid + 1;
}
}
if(matrix[r / m][r % m] != target) return false;
return true;
}
}
lc35 搜索插入位置https://leetcode.cn/problems/search-insert-position/description/
public int searchInsert(int[] nums, int target) {
//O(logn)很明显的二分
int l = 0,r = nums.length;
while(l < r){
int mid = (l + r) >> 1;
if(nums[mid] >= target){
r = mid;
}else{
l = mid + 1;
}
}
return r;
}
lc34. 在排序数组中查找元素的第一个和最后一个位置https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/description/
class Solution {
public int[] searchRange(int[] nums, int target) {
//O(logn)二分
//两个模板一起用
if(nums.length == 0 || nums == null) return new int[]{
-1,-1};
int l = 0, r = nums.length - 1;
//先找start
while(l < r){
int mid = (l + r) >> 1;
if(nums[mid] >= target){
r = mid;
}else{
l = mid + 1;
}
}
if(nums[r] != target) return new int[]{
-1,-1};
int start = r;
//找end
l = 0;
r = nums.length - 1;
while(l < r){
int mid = (l + r + 1) >> 1;
if(nums[mid] <= target){
l = mid;
}else{
r = mid - 1;
}
}
int end = l;
return new int[]{
start,end};
}
}
lc153. 寻找旋转排序数组中的最小值https://leetcode.cn/problems/find-minimum-in-rotated-sorted-array/description/
class Solution {
public int findMin(int[] nums) {
//找一个状态能够二段性
int n = nums.length;
int l = 0, r = n - 1;
while(l < r){
int mid = (l + r) >> 1;
if(nums[mid] < nums[n - 1]){
r = mid;
}else{
l = mid + 1;
}
}
return nums[l];
}
}
lc33 搜索旋转排序数组https://leetcode.cn/problems/search-in-rotated-sorted-array/description/
class Solution {
public int search(int[] nums, int target) {
//先找出最小值的下标
int minIndex = findMinIndex(nums);
System.out.println(minIndex);
if(target <= nums[nums.length - 1]){
//后半段
return binarySearch(nums,target,minIndex,nums.length - 1);
}else{
//前半段
return binarySearch(nums,target,0,minIndex - 1);
}
}
public int binarySearch(int[] nums, int target, int left, int right){
int l = left, r = right;
while(l < r){
int mid = (l + r) >> 1;
if(nums[mid] >= target){
r = mid;
}else{
l = mid + 1;
}
}
System.out.println(l);
if(nums[l] == target) return l;
return -1;
}
public int findMinIndex(int[] nums) {
//找一个状态能够二段性
int n = nums.length;
int l = 0, r = n - 1;
while(l < r){
int mid = (l + r) >> 1;
if(nums[mid] < nums[n - 1]){
r = mid;
}else{
l = mid + 1;
}
}
return l;
}
}
- 总结nums[mid] >= target可以用于找第一个出现的target值 不存在则返回-1
- nums[mid] <= target可以用于找最后一个出现的target值 不存在则返回-1