前言：二分查找的思路简单，但是细节需要格外注意
二分查找
先看以下两份模板：
有序数组找到目标数并返回下标，没有就返回-1
1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
| public int search(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = (left + right) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] < target) {
left = mid + 1;
} else if (nums[mid] > target) {
right = mid - 1;
}
}
return -1;
}
|
2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
| class Solution {
public int search(int[] nums, int target) {
int left = 0;
int right = nums.length;
while (left < right) {
int mid = (left + right) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] < target) {
left = mid + 1;
} else if (nums[mid] > target) {
right = mid;
}
}
return -1;
}
}
|
上面两份代码的区别在于3处不同
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
| int right = nums.length;
或者
int right = nums.length -1;
while (left < right) {
}
或者
while (left <= right) {
}
right = mid -1;
或者
right = mid;
|
这些不同在于区间
第一份代码区间为[left,right],左右区间都是闭合，所以right需要取length-1,循环条件只要left<=right就继续运行，right = mid - 1，因为mid值已经不等于target了。
第二份代码区间为[left,right),分析略。
当然以上的代码只适用只有一个目标值的情况
寻找左侧边界的二分搜索
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
| int left_bound(int[] nums, int target) {
if (nums.length == 0) {
return -1;
}
int left = 0;
int right = nums.length;
while (left < right) {
int mid = (left + right) / 2;
if (nums[mid] == target) {
right = mid;
} else if (nums[mid] < target) {
left = mid + 1;
} else if (nums[mid] > target) {
right = mid;
}
}
if (left == nums.length) {
return -1;
}
return nums[left] == target ? left : -1;
}
|
区间为[left,right)
寻找右侧边界的二分搜索
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
| int right_bound(int[] nums, int target) {
if (nums.length == 0) {
return -1;
}
int left = 0, right = nums.length;
while (left < right) {
int mid = (left + right) / 2;
if (nums[mid] == target) {
left = mid + 1;
} else if (nums[mid] < target) {
left = mid + 1;
} else if (nums[mid] > target) {
right = mid;
}
}
if (left == 0) {
return -1;
}
return nums[left - 1] == target ? (left - 1) : -1;
}
|