- 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) { right = mid - 1; } else if(nums[mid] == target) { // 直接返回 return mid; } } // 直接返回 return -1; } int left_bound(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) { right = mid - 1; } else if (nums[mid] == target) { // 别返回,收缩左侧边界 right = mid - 1; } } // 最后要检查 left 越界的情况 if (left >= nums.length || nums[left] != target) return -1; return left; } int right_bound(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) { right = mid - 1; } else if (nums[mid] == target) { // 别返回,收缩右侧边界 left = mid + 1; } } // 最后要检查 right 越界的情况 if (right < 0 || nums[right] != target) return -1; return right; }