[array] leetcode - 34. Search for a Range - Medium
leetcode - 34. Search for a Range - Medium
descrition
Given an array of integers sorted in ascending order, 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].
解析
对于有序数组的查找问题,基本上都可以使用折半查找的思路。再者题目的要求复杂度是 O(log n),因此我们需要两次折半查找。如果我们只用一次折半查找,找到数组 target 出现的任意一个位置,然后线性遍历找到最左边和最右,需要的复杂度是 O(n)。
具体实现代码如下。
code
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution{
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> ans(2, -1);
int ileft = binarySearchLeftMost(nums, target);
if(ileft < 0)
return ans;
int iright = binarySearchRightMost(nums, target);
ans[0] = ileft;
ans[1] = iright;
return ans;
}
int binarySearchLeftMost(vector<int>& nums, int target){
int ans = -1; // save the left most index in nums which equal to target
int ileft = 0, iright = nums.size() - 1;
while(ileft <= iright){
int imid = ileft + (iright - ileft) / 2;
if(nums[imid] == target){
ans = imid;
iright = imid - 1;
}else if (nums[imid] < target){
ileft = imid + 1;
}else{
// nums[imid] > target
iright = imid - 1;
}
}
return ans;
}
int binarySearchRightMost(vector<int>& nums, int target){
int ans = -1;
int ileft = 0, iright = nums.size() - 1;
while(ileft <= iright){
int imid = ileft + (iright - ileft) / 2;
if(nums[imid] == target){
ans = imid;
ileft = imid + 1;
}else if (nums[imid] < target){
ileft = imid + 1;
}else {
// nums[imid] > target
iright = imid - 1;
}
}
return ans;
}
};
int main()
{
return 0;
}
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