34. Search for a Range

一句话思路：新建数组存储位置，输出数组

一刷报错：

是nums[mid]，nums[start/end]和target的关系，不是mid start end和target的关系
右边界 start end调换后，不对

二刷改了小格式之后ac：

数组定义时，举例出元素的时候没有等号。return new int[] {-1, -1};
是start end=bound[0,1]，不是target＝bound[0,1]。别粗心。

总结：bound[0]，是从前往后查找的，从前开始查。

bound[1]，是从后往前查找的，从后开始查。

风格上：

数组里面的数之间要有空格。
每次声明一个变量 int a;

　　　　　　　　　　　　intb;

if 后面要加空格！！

class Solution {public int[] searchRange(int[] nums, int target) {if (nums.length == 0) {return new int[]{-1, -1};}int start;int end;int mid;int []bound = new int[2];//search for the left boundstart = 0;end = nums.length - 1;while (start + 1 < end) {mid = start + (end - start) / 2;if (nums[mid] == target) {end = mid;}if (nums[mid] < target) {start = mid;}if (nums[mid] > target) {end = mid;}}//System.out.println("start1="+start);//System.out.println("end1="+end);if (nums[start] == target) {bound[0] = start;}else if (nums[end] == target) {bound[0] = end;}else {bound[0] = -1;bound[1] = -1;}//System.out.println("bound[0]="+bound[0]);//search for the right boundend = 0;start = nums.length - 1;while (end + 1 < start) {mid = end + (start - end) / 2;if (nums[mid] == target) {end = mid;}if (nums[mid] < target) {end = mid;}if (nums[mid] > target) {start = mid;}}//System.out.println("start2="+start);//System.out.println("end2="+end);if (nums[start] == target) {bound[1] = start;}else if (nums[end] == target) {bound[1] = end;}else {bound[0] = -1;bound[1] = -1;}//System.out.println("bound[1]="+bound[1]);return bound;}
}

转载于:https://www.cnblogs.com/immiao0319/p/7877497.html

34. Search for a Range相关推荐

LeetCode - 34. Search for a Range
34. Search for a Range Problem's Link -------------------------------------------------------------- ...
LeetCode 34. Search for a Range
LeetCode 34. Search for a Range Solution1:我的答案竟然超过了100%,啊哈哈哈哈哈 class Solution { public:vector<in ...
[LeetCode] Search for a Range [34]
题目 Given a sorted array of integers, find the starting and ending position of a given target value. ...
二分法：查找区间search for a range
问题描述: 给定一个排序数组nums(nums中有重复元素)与目标值target,如果 target在nums里出现,则返回target所在区间的左右端点下标,[左端点, 右端点],如果target在 ...
【leetcode】Search for a Range
题目描述: Given a sorted array of integers, find the starting and ending position of a given target valu ...
UnicodeDecodeError: 'ascii' codec can't decode byte 0xe2 in position 34: ordinal not in range(128)
以上错误是我在使用pip 安装 pyvex时候出现系统 ubuntu14.04 32位解决办法: vim ~/.bashrc(打开用户配置文件) 在文件最后一行添加 export LC_ALL= ...
Lintcode61 Search for a Range solution 题解
[题目描述] Given a sorted array of n integers, find the starting and ending position of a given target v ...
继续过中等难度.0309
. 8 String to Integer (atoi) 13.9% Medium . 151 Reverse Words in a String 15.7% Mediu ...
Leetcode重点250题
LeetCode重点250题这个重点题目是把LeetCode前400题进行精简.精简方法如下: 删除不常考,面试低频出现题目删除重复代码题目(例:链表反转206题,代码在234题出现过) 删除过于 ...

34. Search for a Range

34. Search for a Range相关推荐

最新文章

热门文章