题解 | #二分查找-I#
二分查找-I
https://www.nowcoder.com/practice/d3df40bd23594118b57554129cadf47b
class Solution { public: /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param nums int整型vector * @param target int整型 * @return int整型 */ int search(vector<int>& nums, int target) { // write code here int tot = nums.size(); int start = 0; if (nums.size() == 0) return -1; if (target == nums[0]) return 0; if (target ==nums[tot - 1]) return (tot - 1); if (nums[tot - 1] == nums[0]) return -1; if (tot - start < 2) return -1; // int idx = tot * (target - nums[0]) / (nums[tot - 1] - nums[0]); int idx = (start + tot) >> 1; while (start < tot) { if (nums[idx] == target) return idx; idx = (idx + tot) >> 1; if (nums[idx] < target) { start = idx + 1; } else { tot = idx - 1; } } return -1; } };