二分查找模板（查找数、左边界、右边界）
Although the basic idea of binary search is comparatively straightforward, the details can be surprisingly tricky...
二分查找框架
int binarySearch(int[] nums, int val) {
int left = 0, right = ...;
while(...) {
int mid = left + (right - left) / 2;
if(nums[mid] == val) {
//...
}else if(nums[mid] < val) {
//...
}else if(nums[mid] > val) {
//...
}
}
return ...;
}
查找一个数val
int binarySearch(int[] nums, int val) {
int left = 0;
int right = nums.length - 1; //☚☚☚
while(left <= right) {
//☚☚☚
int mid = left + (right - left) / 2;
if(nums[mid] == val) {
return mid; //☚☚☚
}else if(nums[mid] < val) {
left = mid + 1;
}else if(nums[mid] > val) {
right = mid - 1;
}
}
return -1;
}
此算法的缺陷
- 若要在
[1, 2, 2, 2, 3]中查找2，则返回的下标为2 - 若为第一个则是下标
1 - 若为最后一个则是下标
3
查找一个数val（返回第一次出现的下标）
方法一：
int binarySearch(int[] nums, int val) {
if(nums == null || nums.length == 0) return -1;
int left = 0;
int right = nums.length - 1; //☚☚☚
while(left <= right) {
//☚☚☚
int mid = left + (right - left) / 2;
if(nums[mid] == val) {
right = mid - 1; //☚☚☚
}else if(nums[mid] < val) {
left = mid + 1;
}else if(nums[mid] > val) {
right = mid - 1;
}
}
return right + 1; //☚☚☚
}
简化代码为
...
while(left <= right) {
int mid = left + (right - left) / 2;
if(nums[mid] < val) {
left = mid + 1;
}else {
right = mid - 1;
}
}
...
方法二：
int binarySearch(int[] nums, int val) {
if(nums == null || nums.length == 0) return -1;
int left = 0;
int right = nums.length; //☚☚☚
while(left < right) {
int mid = left + (right - left) / 2;
if(nums[mid] == val) {
right = mid; //☚☚☚
}else if(nums[mid] < val){
left = mid + 1;
}else if(nums[mid] > val){
right = mid; //☚☚☚
}
}
return left; //☚☚☚
}
查找一个数val（返回最后一次出现的下标）
方法一：
int binarySearch(int[] nums, int val) {
if(nums == null || nums.length == 0) return -1;
int left = 0;
int right = nums.length - 1; //☚☚☚
while(left <= right) {
//☚☚☚
int mid = left + (right - left) / 2;
if(nums[mid] == val) {
left = mid + 1; //☚☚☚
}else if(nums[mid] < val) {
left = mid + 1;
}else if(nums[mid] > val) {
right = mid - 1;
}
}
return left - 1; //☚☚☚
}
简化代码为
...
while(left <= right) {
int mid = left + (right - left) / 2;
if(nums[mid] <= val) {
left = mid + 1;
}else {
right = mid - 1;
}
}
...
方法二：
int binarySearch(int[] nums, int val) {
if(nums == null || nums.length == 0) return -1;
int left = 0;
int right = nums.length; //☚☚☚
while(left < right) {
//☚☚☚
int mid = left + (right - left) / 2;
if(nums[mid] == val) {
left = mid + 1; //☚☚☚
}else if(nums[mid] < val){
left = mid + 1;
}else if(nums[mid] > val){
right = mid; //☚☚☚
}
}
return left - 1; //☚☚☚
}
左边界（返回小于val的最大值的下标）
同查找一个数（返回第一次出现的下标）思路一样，
找到第一次出现的下标right + 1后，下标right处就是小于它的最大值
int binarySearch(int[] nums, int val) {
if(nums == null || nums.length == 0) return -1;
int left = 0;
int right = nums.length - 1;
while(left <= right) {
int mid = left + (right - left) / 2;
if(nums[mid] == val) {
right = mid - 1;
}else if(nums[mid] < val) {
left = mid + 1;
}else if(nums[mid] > val) {
right = mid - 1;
}
}
return right; //☚☚☚
}
右边界（返回大于val的最小值的下标）
同查找一个数（返回最后一次出现的下标）思路一样，
找到最后一次出现的下标left - 1后，下标left处就是大于它的最小值
int binarySearch(int[] nums, int val) {
if(nums == null || nums.length == 0) return -1;
int left = 0;
int right = nums.length - 1;
while(left <= right) {
int mid = left + (right - left) / 2;
if(nums[mid] == val) {
left = mid + 1;
}else if(nums[mid] < val) {
left = mid + 1;
}else if(nums[mid] > val) {
right = mid - 1;
}
}
return left; //☚☚☚
}
上述代码while循环中分情况写全是为了便于理解，看懂就好，是否化简看个人喜好啦~
相关参考链接：
labuladong二分查找算法详解：https://mp.weixin.qq.com/s/uA2suoVykENmCQcKFMOSuQ
二分模板大全：https://blog.csdn.net/vocaloid01/article/details/82257336