LEET code --Search for a Range

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].

直觀就是直接找一次就好
一開始寫玩都不會對
搞半天

*returnSize 要給2

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int* searchRange(int* nums, int numsSize, int target, int* returnSize) { *returnSize = 2; int *ans = (int *)malloc(sizeof(int) * *returnSize); ans[0]=-1; ans[1]=-1; if(numsSize == 0)return ans; if(target < nums[0] || target > nums[numsSize-1])return ans; for(int i=0;i<numsSize;i++){ if(target == nums[i]){ if(ans[0]==-1){ ans[0]=i; ans[1]=i; } else ans[1]=i; } if(target<nums[i])break; } return ans; }

如果要更快可以改用binary search
因為都排序好了
會兩倍快