LeetCode c语言-Search for a Range
Title:
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]
.
这道题给定一个排序好的数组,然后找到重复的数字,并输出序号即可。
该题的要求是时间复杂度为O(log n),也就是说不能采用从头遍历到尾的方法,容易超时。因此采用另一种做法,从两边同时向中间靠拢。由于是排序好的数组,因此判断逻辑较为简单。
Solution:
int* searchRange(int* nums, int numsSize, int target, int* returnSize) {
int *result=(int*)malloc(sizeof(int)*2);
int i,j;
int tmp;
if (numsSize<=0) {
*returnSize=2;
result[0]=-1;
result[1]=-1;
return result;
}
if (numsSize==1) {
if (target==nums[0]) {
*returnSize=2;
result[0]=0;
result[1]=0;
return result;
}
else {
*returnSize=2;
result[0]=-1;
result[1]=-1;
return result;
}
}
i=0;
j=numsSize-1;
while(i<j) {
if (target<nums[i] || target>nums[j]){
*returnSize=2;
result[0]=-1;
result[1]=-1;
return result;
}
else if (target==nums[i]) {
tmp=i+1;
while(tmp<numsSize && nums[tmp]==nums[i]) {
tmp++;
}
*returnSize=2;
result[0]=i;
result[1]=tmp-1;
return result;
}
else if (target==nums[j]) {
tmp=j-1;
while(tmp>=0 && nums[tmp]==nums[j]) {
tmp--;
}
*returnSize=2;
result[0]=tmp+1;
result[1]=j;
return result;
}
else {
if (i+1==j-1)
i++;
else {
i++;
j--;
}
}
}
*returnSize=2;
result[0]=-1;
result[1]=-1;
return result;
}
posted on 2017-09-26 10:30 sichenzhao 阅读(117) 评论(0) 编辑 收藏 举报