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34_Search_for_a_Range

question

34. Search for a Range

Original Page

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].

analysis

一开始就想到了二分查找,但是原来做二分查找的时候一般都是找到确定的那个数就完成了,
这里的情况比较特殊,需要找到整个区间,所以需要两遍查找,并且一个是找到小于target
的最大索引,一个是找到大于target的最大索引,代码参考leetcode discuss,这位仁
兄也做了详细的分析解释。

code

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class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> ret(2, -1);
int i = 0, j = nums.size() - 1;
int mid;
while(i < j){
mid = (i + j) / 2;
if(nums[mid] < target) i = mid + 1;
else j = mid;
}
if(nums[i] != target) return ret;
else {
ret[0] = i;
if((i+1) < (nums.size() - 1) && nums[i+1] > target){
ret[1] = i;
return ret;
}
} //一点小优化
j = nums.size() - 1;
while(i < j){
mid = (i + j) / 2 + 1;
if(nums[mid] > target) j = mid - 1;
else i = mid;
}
ret[1] = j;
return ret;
}
};
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