二分查找(LeetCode #704)
二分查找(LeetCode #704)
解题思路
-
基本的二分搜索
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搜索区间为[left, right],所以终止条件为
left<=right
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为了防止溢出,使用
mid = left + (right - left) / 2
代码
class Solution {
public int search(int[] nums, int target) {
int left = 0;
int right = nums.length - 1;
int mid = 0;
while(left <= right) {
mid = left + (right - left) / 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;
}
}