First Occurrence
给定一个排序好的数组和一个target value，找出target第一次出现的index
假设:
数组中可以有重复的元素。
实例:
a = 1, 2, 3 }，t＝2，返回1
a = { 1, 2, 3 }，t＝4，返回- 1
a = { 1, 2, 2，2, 3 }，t＝2，返回1
public class Solution {
public int firstOccur(int[] arr, int target) {
if(arr==null||arr.length==0)return-1;
int left = 0;
int right = arr.length-1;
while(left<right-1){
int mid = left+(right-left)/2;
if(arr[mid]<target){
left = mid;
}else{
right = mid;
}
}
if(arr[left]==target){
return left;
}
if(arr[right]==target){
return right;
}
return -1;
}
}
//物理意义：
//left < target <= right
Last Occurrence
找出target最后一次出现的index，其他条件与上题一致
public class Solution {
public int lastOccur(int[] arr, int target) {
if(arr == null || arr.length==0)return -1;
int left = 0;
int right = arr.length-1;
while(left<right-1){
int mid = left+(right-left)/2;
if(arr[mid]<=target){
left = mid;
}else{
right = mid;
}
}
if(arr[right] == target)return right;
if(arr[left] == target)return left;
return -1;
}
}
//物理意义：
//left <= target < right
Closest In Sorted Array, 找到最接近target的index
Given a target integer T and an integer array A sorted in ascending order, find the index i in A such that A[i] is closest to T.
Assumptions:
There can be duplicate elements in the array, and we can return any of the indices with same value.
Examples:
A = {1, 2, 3}, T = 2, return 1
A = {1, 4, 6}, T = 3, return 1
A = {1, 4, 6}, T = 5, return 1 or 2
A = {1, 3, 3, 4}, T = 2, return 0 or 1 or 2
public class Solution {
public int closest(int[] arr, int target) {
if(arr==null||arr.length==0)return -1;
int left = 0;
int right = arr.length-1;
while(left < right-1){
int mid = left+(right-left)/2;
if(arr[mid]==target)return mid;
if(arr[mid]<target){
left = mid;
}else{
right = mid;
}
}
int abs1 = Math.abs(arr[left]-target);
int abs2 = Math.abs(arr[right]-target);
return abs1>abs2?right:left;
}
}
//物理意义：
//left <= target <= right
Search In Sorted Matrix I
给定一个只包含整数的二维矩阵，每一行按升序排序。下一行的第一个元素大于（或等于）上一行的最后一个元素。
给定一个target，返回target位于矩阵中的index。如果target number不存在于矩阵中，返回{ - 1，- 1 }。
假设:
给定的矩阵不是空的，并且有n×m的大小，其中n>＝0和m>＝0。
实例：
矩阵= { {1, 2, 3 }，{ 4, 5, 7 }，{ 8, 9, 10 } }
目标= 7，返回{ 1, 2 }
目标= 6，返回{ - 1，- 1 }代表矩阵中不存在的目标数。
public class Solution {
public int[] search(int[][] matrix, int target) {
int[] res = {-1,-1};
int left = 0;
int m = matrix.length;
int n = matrix[0].length;
int right = m*n-1;
while(left<=right){
int mid = left+(right-left)/2;
if(matrix[mid/n][mid%n] == target){
res[0] = mid/n;
res[1] = mid%n;
return res;
}
if(target<matrix[mid/n][mid%n]){
right = mid-1;
}else{
left = mid+1;
}
}
return res;
}
}
K Closest In Sorted Array
Given a target integer T, a non-negative integer K and an integer array A sorted in ascending order, find the K closest numbers to T in A.
Assumptions
A is not null
K is guranteed to be >= 0 and K is guranteed to be <= A.length
Return
A size K integer array containing the K closest numbers(not indices) in A, sorted in ascending order by the difference between the number and T.
Examples
A = {1, 2, 3}, T = 2, K = 3, return {2, 1, 3} or {2, 3, 1}
A = {1, 4, 6, 8}, T = 3, K = 3, return {4, 1, 6}
public class Solution {
//time:O(lgn+k); space:O(k);
public int[] kClosest(int[] arr, int target, int k) {
if(arr == null || arr.length == 0){
return arr;
}
if( k == 0 ){
return new int[0];
}
int[] result = new int[k];
int left = largestSmallerEqual(arr, target);
int right = left + 1;
for(int i=0; i<k; i++){//O(k)
if(right>=arr.length || left>=0 && target-arr[left]<=arr[right]-target){
result[i] = arr[left--];
}else{
result[i] = arr[right++];
}
}
return result;
}
//O(lgn)
private int largestSmallerEqual(int[] arr, int target){
int left = 0;
int right = arr.length-1;
while ( left<right-1 ){//left<=target<right
int mid = left+(right-left)/2;
if(arr[mid]<=target){
left = mid;
}else{
right = mid;
}
}
if(arr[right]<=target){
return right;
}
if(arr[left]<=target){
return left;
}
return -1;
}
}
