给定数组需是有序数组。 二分法查找的时间复杂度为logn。 int binary_search(int[] nums, int target) { int left = 0, right = nums.length - 1; while(left <= right) { int mid = left + (right - left) / 2; if (nums[mid] < target) { left = mid + 1; } else if (nums[mid] > target)...