版权声明：be the one ~you will be the one~~ https://blog.csdn.net/Hqxcsdn/article/details/87923244
34. 在排序数组中查找元素的第一个和最后一个位置
采用二分查找 ，轻松查出
/**
*
*/
package test;
/***
* @author 18071
* @Date 2019年2月25日 功能:
***/
public class test {
public static void main(String args[]) {
int[] nums = { 5, 5, 5, 5, 5, 5, 5, 7, 7, 8, 8, 10 };
int target = 8;
Solution s = new Solution();
if (s.find(nums, target, 0, 11) != null) {
System.out.println("w2222ow");
} else {
}
}
}
class remm {
public int t(int x) {
if (x > 5) {
return 5;
} else if (x < 5) {
}
return 123;
}
}
class Solution {
public int[] find(int[] nums, int target, int low, int heigh) {
int mid;
mid = (low + heigh + 1) / 2;
int templow=0;
int tempheigh=0;
if (target == nums[mid]) {
templow = mid;
tempheigh = mid;
while (templow > 0) {
templow--;
if (nums[templow] != target) {
templow++;
System.out.println(templow + "is templow " + nums[templow]);
break;
}
}
while (tempheigh < heigh) {
System.out.println(tempheigh + "is tempheigh " + nums[tempheigh]);
tempheigh++;
if (nums[tempheigh] != target) {
tempheigh--;
//
break;
}
}
System.out.println("100 行 " + templow + " " + tempheigh);
int[] rua = { templow, tempheigh };
for (int i = 0; i < 2; i++) {
System.out.println(rua[i]);
}
return rua;
}
if (target < nums[mid] && target > nums[low]) {
heigh = mid;
return find(nums, target, low, heigh);
}
if (target > nums[mid] && target < nums[heigh]) {
low = mid;
return find(nums, target, low, heigh);
}
return null;
}
}