LeetCode——Search for a Range
Description:
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].
public class Solution {
public int[] searchRange(int[] nums, int target) {
int flag = -1, start=-1, end = -1;
for(int i=0; i<nums.length; i++) {
if(nums[i] == target) {
start = i;
for(int j=i; j<nums.length; j++) {
if(nums[j] != target) {
end = j-1;
flag = 1;
break;
}
if(nums[j]==target && j==nums.length-1) {
end = j;
flag = 1;
break;
}
}
}
if(flag == 1) {
break;
}
}
return new int[]{start, end};
}
}
作者:Pickle
声明:对于转载分享我是没有意见的,出于对博客园社区和作者的尊重一定要保留原文地址哈。
致读者:坚持写博客不容易,写高质量博客更难,我也在不断的学习和进步,希望和所有同路人一道用技术来改变生活。觉得有点用就点个赞哈。








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