class Solution {
public:
// 缩小搜索空间法
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> res(2, -1);
if(nums.size() == 0){
return res;
}
int left = 0;
int right = nums.size() - 1;
// 查找第一个位置
while(left < right){
int mid = left + (right - left) / 2;
if(nums[mid] < target){
left = mid+1;
}
else{
right = mid;
}
}
if(nums[left] == target){
res[0] = left;
}
left = 0;
right = nums.size() - 1;
while(left < right){
int mid = left + (right - left+1) / 2;
if(nums[mid] > target){
right = mid - 1;
}
else{
left = mid;
}
}
if(nums[left] == target){
res[1] = left;
}
return res;
}
};
思路依然是逐步缩小 搜索空间。这里主要是一个左中位数和右中位数的概念。

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