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Commit368b74e

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author
applewjg
committed
Search in Rotated Sorted Array
Change-Id: I593d375bd9c6c383e51595186daa86c09df2c275
1 parentc4b11fb commit368b74e

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2 files changed

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‎SearchinRotatedSortedArray.java

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/*
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Author: King, wangjingui@outlook.com
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Date: Dec 20, 2014
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Problem: Search in Rotated Sorted Array
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Difficulty: Medium
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Source: https://oj.leetcode.com/problems/search-in-rotated-sorted-array/
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Notes:
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Suppose a sorted array is rotated at some pivot unknown to you beforehand.
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(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
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You are given a target value to search. If found in the array return its index, otherwise return -1.
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You may assume no duplicate exists in the array.
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Solution: Binary search. O(lgn) eg. [4 5 6] -7- 8 1 2, 5 6 0 -1- [2 3 4]
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*/
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publicclassSolution {
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publicintsearch(int[]A,inttarget) {
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inti =0,j =A.length -1;
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while (i <=j) {
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intmid = (i +j) /2;
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if (A[mid] ==target)
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returnmid;
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if (A[i] <=A[mid]) {
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if (A[i] <=target &&target <A[mid])
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j =mid -1;
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else
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i =mid +1;
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}else {
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if (A[mid] <target &&target <=A[j])
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i =mid +1;
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else
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j =mid -1;
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}
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}
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return -1;
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}
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}

‎SearchinRotatedSortedArrayII.java

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/*
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Author: King, higuige@gmail.com
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Date: Nov 18, 2014
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Problem: Search in Rotated Sorted Array II
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Difficulty: Medium
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Source: https://oj.leetcode.com/problems/search-in-rotated-sorted-array-ii/
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Notes:
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Suppose a sorted array is rotated at some pivot unknown to you beforehand.
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(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
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What if duplicates are allowed?
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Would this affect the run-time complexity? How and why?
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Write a function to determine if a given target is in the array.
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Solution: Sequence search. O(n)
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Since there are duplicates, it's hard to decide which branch to go if binary-search is deployed.
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*/
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publicclassSolution {
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publicbooleansearch_1(int[]A,inttarget) {
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for (inti =0;i <A.length; ++i) {
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if (A[i] ==target)returntrue;
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}
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returnfalse;
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}
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publicbooleansearch(int[]A,inttarget) {
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intleft =0,right =A.length -1;
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while (left <=right) {
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intmid = (left +right) /2;
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if (A[mid] ==target)returntrue;
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if (A[left] <A[mid]) {
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if (A[left] <=target &&target <A[mid])right =mid -1;
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elseleft =mid +1;
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}elseif (A[left] >A[mid]) {
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if (A[mid] <target &&target <=A[right])left =mid +1;
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elseright =mid -1;
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}else {
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if (A[left] ==A[right]) {
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++left; --right;
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}elseleft =mid +1;
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}
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}
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returnfalse;
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}
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}

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