leetcode做题总结，题目Search for a Range 2012/03/02

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题目是求出一个有序列表有多个重复的target的第一个和最后一个的index，我的做法是先用二分找出其中一个target，然后向两边推进找出最前和最后的位置。

 public int[] searchRange(int[] A, int target) {        int[] res ={-1,-1};        if(A.length==0)return res;        int i=0, j=A.length-1, p=-1;        while(i<=j){            int m=(i+j)/2;            if(A[m]==target){                p=m;                break;            }             if(A[m]<target) i=m+1;            else j=m-1;        }        if(p==-1)return res;        i=p;        j=p;        while(i>=0&&A[i]==target){            i--;        }        while(j<=A.length-1&&A[j]==target){            j++;        }        res[0]=i+1;        res[1]=j-1;        return res;    }

Update 2015/08/22: 上面的解法不对，不是logn，正确的解法是使用两次二分法找到对应的左、右节点。

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 search(ArrayList<Integer> A, int start, int end, boolean left, int target){        while (start <= end){            int mid = (start + end) / 2;            if (A.get(mid) == target){                if (left){                    if (mid == 0 || A.get(mid) > A.get(mid - 1)){                        return mid;                    } else {                        end = mid - 1;                    }                } else {                    if (mid == A.size() - 1 || A.get(mid) < A.get(mid + 1)){                        return mid;                    } else {                        start = mid+1;                    }                }            } else if (A.get(mid) < target){                start = mid+1;            } else {                end = mid-1;            }        }        return -1;            }        public ArrayList<Integer> searchRange(ArrayList<Integer> A, int target) {        // write your code here        ArrayList<Integer> res = new ArrayList<Integer>();        res.add(search(A, 0, A.size() - 1, true, target));        res.add(search(A, 0, A.size() - 1, false, target));        return res;    }}

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