【LeetCode】Search For A Range 区间搜索[二分法]

给定一个包含 n 个整数的排序数组，找出给定目标值 target 的起始和结束位置。

如果目标值不在数组中，则返回[-1, -1]

样例
给出[5, 7, 7, 8, 8, 10]和目标值target=8,
返回[3, 4]

挑战
时间复杂度 O(log n)

标签
二分法 数组 排序数组
相关题目
中等 Range Sum Query 2D - Immutable 30 %

---

SOLUTION 1:
使用改进的二分查找法。终止条件是：left + 1 < right 这样结束的时候，会有2个值供我们判断。这样做的最大的好处是，不用处理各种越界问题。

1. 先找左边界。当mid == target,将right移动到mid，继续查找左边界。
   最后如果没有找到target,退出
2. 再找右边界。当mid == target,将left移动到mid，继续查找右边界。
   最后如果没有找到target,退出
   （1）Java

public class Solution {    /*     * @param A: an integer sorted array     * @param target: an integer to be inserted     * @return: a list of length 2, [index1, index2]     */    public int[] searchRange(int[] A, int target) {        if(A.length == 0){            return new int[]{-1,-1};        }        int start, end, mid;        int[] bound = new int[2];        //因为是排序数组，无需考虑target中夹杂某数的情况        //search for left bound:        start = 0;        end = A.length - 1;        while(start + 1 < end){            mid = start + (end - start) / 2;            if(A[mid] == target){                end = mid;            }else if(A[mid] < target){                start = mid;            }else{ //if (A[mid] > target)                end = mid;            }        }        if(A[start] == target){            bound[0] = start;        }else if(A[end] == target){            bound[0] = end;        }else{            bound[0] = bound[1] = -1;            return bound;        }        //search for ringht bound        start = 0;        end = A.length - 1;        while(start + 1 < end){            mid = start + (end - start) / 2;            if(A[mid] == target){                start = mid;            }else if(A[mid] < target){                start = mid;            }else{ //if (A[mid] > target)                end = mid;            }        }        if(A[end] == target){            bound[1] = end;        }else if(A[start] == target){            bound[1] = start;        }else{            bound[0] = bound[1] = -1;            return bound;        }        return bound;    }}

(2)C++

class Solution {  /**    *@param A : an integer sorted array   *@param target :  an integer to be inserted   *return : a list of length 2, [index1, index2]   */public:  vector<int> searchRange(vector<int> &A, int target) {    // write your code here    vector<int> ans;    int ansl = -1;    for (int l = 0, r = A.size() - 1; l <= r;) {      int mid = l + (r - l) / 2;      if (A[mid] > target) {        r = mid - 1;      }      if (A[mid] < target) {        l = mid + 1;      }      if (A[mid] == target) {        ansl = mid;        r = mid - 1;      }    }    int ansr = -1;    for (int l = 0, r = A.size() - 1; l <= r;) {      int mid = l + (r - l) / 2;      if (A[mid] > target) {        r = mid - 1;      }      if (A[mid] < target) {        l = mid + 1;      }      if (A[mid] == target) {        ansr = mid;        l = mid + 1;      }    }    ans.push_back(ansl);    ans.push_back(ansr);    return ans;  }};