LeetCode 34. Search for a Range（搜索范围）

原题网址：https://leetcode.com/problems/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]

.
方法：二分法查找。

public class Solution {
public int[] searchRange(int[] nums, int target) {
int[] range = new int[2];
range[0] = -1;
range[1] = -1;
int i=0, j=nums.length-1;
while (i<=j) {
int m = (i+j)/2;
if (target == nums[m]) {
range[0] = m;
j = m-1;
} else if (target < nums[m]) {
j = m - 1;
} else {
i = m + 1;
}
}
i=0;
j=nums.length-1;
while (i<=j) {
int m = (i+j)/2;
if (target == nums[m]) {
range[1] = m;
i = m + 1;
} else if (target < nums[m]) {
j = m - 1;
} else {
i = m + 1;
}
}
return range;
}
}