Search for a Range
Search for a Range
问题：
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].
思路：
递归二分查找问题
我的代码：
public class Solution { public int[] searchRange(int[] A, int target) { int[] rst = {-1,-1}; if(A == null || A.length == 0) return rst; int left = 0; int right = A.length - 1; return helper(A, target, left, right); } public int[] helper(int[] A, int target, int left, int right) { int[] rst = {-1,-1}; if(left > right) return rst; if(left == right) { if(target == A[left]) { rst[0] = left; rst[1] = left; return rst; } return rst; } int mid = (left + right)/2; if(A[mid] == target) { rst[0] = mid; rst[1] = mid; int[] leftRange = helper(A, target, left, mid - 1); int[] rightRange = helper(A, target, mid + 1, right); return mergeRange(leftRange, rightRange, rst); } else if(A[mid] > target) return helper(A, target, left, mid - 1); else return helper(A, target, mid + 1, right); } public int[] mergeRange(int[] left, int[]right, int[] mid) { int[] rst = new int[2]; int min = Integer.MAX_VALUE; int max = Integer.MIN_VALUE; for(int num: left) { if(num >= 0) { min = Math.min(min, num); max = Math.max(max, num); } } for(int num: right) { if(num >= 0) { min = Math.min(min, num); max = Math.max(max, num); } } for(int num: mid) { if(num >= 0) { min = Math.min(min, num); max = Math.max(max, num); } } rst[0] = min; rst[1] = max; return rst; } }
他人代码：
public class Solution { public int[] searchRange(int[] A, int target) { int start, end, mid; int[] bound = new int[2]; // 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 { 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 right 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 { 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; } }
学习之处：
- 自己的代码虽然AC了，但是判断太多了，一点也不简洁，这为以后的调试，改代码带来了痛苦，而且代码越多越容易出bug啊
- 看看别人的代码，被人代码的模板之前也整理过，但是就没想到用呢，首先left + 1 <right 这种判断的存在是因为 之后的二分存在 left = mid or right = mid 而不是left = mid - 1 right = mid + 1 如果不是Left + 1 < right 则存在死循环
- 对于不知道左边还是右边的：首先考虑可不可以排除一边，如通过left mid right值三者之间的大小关系，如果不能排除，更加普世的作用是使用递归分别dfsleft dfsright进行结合，其实也可以在循环里面做了，方法是首先判断左边，然后判断右边，代码就如同上面别人的代码一样，真实巧妙！