[LeetCode] Search for a Range (sorted integers array,find start & end position of a target number)

/**
O(n)的解法，从左到右遍历。
**/
public class Solution {
    public int[] searchRange(int[] A, int target) {
        if(A.length == 0) {
            return new int[]{-1,-1};
        }
        int i = -1, j = -1;
        for (int k = 0; k < A.length; k++) {
            if(A[k] == target) {
                if (i == j && j == -1){ //first time found target
                    i = k;
                    j = k;
                } else {//second time or later
                    j = k;
                }
            }
        }
        return new int[]{i,j};
    }
}

解法2， O(lgN)的解法，二分法

public int[] searchRange(int[] A, int target) {
    if (A == null || A.length == 0) return new int[]{-1, -1};
    int left = findLeft(A, target);
    int right = findRight(A, target);
    return new int[]{left, right};
}

private int findLeft(int[] A, int target) {
    if (target < A[0] || target > A[A.length-1]) return -1;
    if (target == A[0]) return 0;
    int start = 0, end = A.length-1;
    while (start < end) {
      int mid = start + (end - start) >> 1;
      if (A[mid] == target && A[mid-1] != target) return mid;
      else if (A[mid] < target) start = mid + 1;//go right
      else end = mid - 1;
    }
    return -1;
}

private int findRight(int[] A, int target) {
    if (target < A[0] || target > A[A.length-1]) return -1;
    if (target == A[A.length-1]) return A.length-1;
    int start = 0, end = A.length-1;
    while (start < end) {
      int mid = start + (end - start) >> 1;
      if (A[mid] == target && A[mid+1] != target) return mid;
      else if (A[mid] > target) end = mid - 1; //go left
      else start = mid + 1;
    }
    return -1;
}