Leetcode 34. 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].

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解题思路：

题意转换， 变成找first bound of target and right bound of target.

找first position:

相同的时候砍掉右边

找last position:

相同的时候砍掉左边

题目可以变化问题：问8（target number）出现了几次？

答案：right bound - left bound + 1

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Java code:

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        if(nums.length == 0 ){
            return new int[]{-1, -1};
        }
        
        int start, end, mid;
        int[] bound = new int[2];
        
        //search for left bound
        start = 0;
        end = nums.length-1;
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                end = mid;  
            } else if (nums[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if(nums[start] == target){
            bound[0] = start;
        } else if (nums[end] == target) {
            bound[0] = end;
        } else {
            bound[0] = bound[1] = -1;
            return bound;
        }
        
        //search for right bound
        start = 0;
        end = nums.length-1;
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                start = mid;  
            } else if (nums[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if(nums[end] == target) {
            bound[1] = end;
        }else if (nums[start] == target){
            bound[1] = start;
        }else {
            bound[0] = bound[1] = -1;
            return bound;
        }
        
        return bound;
    }
}

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Reference:

1. http://www.jiuzhang.com/solutions/search-for-a-range/