LeetCode: Search for a Range 解题报告

Search for a Range
Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

SOLUTION 1:

使用改进的二分查找法。终止条件是：left < right - 1 这样结束的时候，会有2个值供我们判断。这样做的最大的好处是，不用处理各种越界问题。

感谢黄老师写出这么优秀的算法：http://answer.ninechapter.com/solutions/search-for-a-range/

请同学们一定要记住这个二分法模板，相当好用哦。

1. 先找左边界。当mid == target,将right移动到mid，继续查找左边界。

 最后如果没有找到target,退出

2. 再找右边界。当mid == target,将left移动到mid，继续查找右边界。

 最后如果没有找到target,退出

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int[] ret = {-1, -1};
        
        if (A == null || A.length == 0) {
            return ret;
        }
        
        int len = A.length;
        int left = 0; 
        int right = len - 1;
        
        // so when loop end, there will be 2 elements in the array.
        // search the left bound.
        while (left < right - 1) {
            int mid = left + (right - left) / 2;
            if (target == A[mid]) {
                // 如果相等，继续往左寻找边界
                right = mid;
            } else if (target > A[mid]) {
                // move right;
                left = mid;
            } else {
                right = mid;
            }
        }
        
        if (A[left] == target) {
            ret[0] = left;
        } else if (A[right] == target) {
            ret[0] = right;
        } else {
            return ret;
        }
        
        left = 0; 
        right = len - 1;
        // so when loop end, there will be 2 elements in the array.
        // search the right bound.
        while (left < right - 1) {
            int mid = left + (right - left) / 2;
            if (target == A[mid]) {
                // 如果相等，继续往右寻找右边界
                left = mid;
            } else if (target > A[mid]) {
                // move right;
                left = mid;
            } else {
                right = mid;
            }
        }
        
        if (A[right] == target) {
            ret[1] = right;
        } else if (A[left] == target) {
            ret[1] = left;
        } else {
            return ret;
        }
        
        return ret;
    }
}

 

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/divide2/SearchRange.java