数组实现二分

34. 在排序数组中查找元素的第一个和最后一个位置 - 力扣（LeetCode） (opens new window)

class Solution {
    public int[] searchRange(int[] nums, int target) {
        return new int[]{left_bound(nums,target),right_bound(nums,target)};
    }
    //常规
    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
                right = mid;
            }
        }
        //直接返回
        return -1;
    }
    //搜索左边界
    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，锁定左侧边界
                right = mid - 1;
            }
        }
        //判断 target 是否存在于nums中
        if (left < 0 || left >= nums.length) {
            return -1;
        }
        //判断一下nums[left]是不是target
        return nums[left] == target ? left : -1;
    }
    //搜索右边界
    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，锁定右侧边界
                left = mid + 1;
            }
        }
        //由于while的结束条件是right == left - 1,且现在在求右边界
        //所以用right代替 left - 1更好记住
        if (right < 0 || right >= nums.length) {
            return -1;
        }     
        return nums[right] == target ? right : -1;
    }
}