# Sorted array a, demand the absolute value of its elements the total number of different values

2010-08-29  来源：本站原创  分类：Tech  人气：149

Sorted array a, demand the absolute value of its elements the total number of different values?

Ideas:

2 index, from 0 to 2 elements closest to the two side began to traverse, each time compare the two index elements where the absolute value, minimum value, and compare the final value of 1 if greater than, then count + + and the value is set to last value, then small forward direction of the index plus 1,

Algorithm complexity:

As long as the traverse 1, the complexity is o (n), n is the number of the array,

Memory footprint:

Need two index, 1 A last, 1 count, so except array, the extra memory is fixed occupation, that is o (1)

Code:

```package test;

public class Test {
public static void main(String[] args) {
int[] is = new int[] { -10, -9, -9, -5, -4, -3, -3, -2, -1, 0, 1, 2, 5, 6, 7, 8, 9, 11, 13, 14 };
System.out.println(funOne(is, locateNumInSortedArray(0, is)));
}

/**
*  Get sorted array  , The absolute value of the number of different  , It is assumed that the array has  0,
*  If the array has no other 0 can write methods to find the nearest  0 Number 2 of    Index of, and slightly modified by method   count  Value to initialize  ,
* @param arr
* @param zeroIndex
* @return
*/
public static int funOne(int[] arr, int zeroIndex) {
int plusIndex = zeroIndex + 1, negativeIndex = zeroIndex - 1;
int last = arr[zeroIndex];
int count = 1;
while (negativeIndex >= 0 || plusIndex < arr.length) {
if (negativeIndex >= 0 && plusIndex < arr.length) {// both side continue
// x = small abs(...)
int x1;
int absNeg = Math.abs(arr[negativeIndex]);
if (absNeg > arr[plusIndex]) {
x1 = arr[plusIndex];
plusIndex += 1;

} else if (absNeg < arr[plusIndex]) {
x1 = absNeg;
negativeIndex -= 1;
} else {
x1 = arr[plusIndex];
plusIndex += 1;
negativeIndex -= 1;
}
if (x1 > last) {
count++;
last = x1;
}
} else if (negativeIndex >= 0) { // only negative site continue
int x2 = Math.abs(arr[negativeIndex]);
negativeIndex -= 1;
if (x2 > last) {
count++;
last = x2;
}
} else { // only plus site continue
int x3 = arr[plusIndex];
plusIndex += 1;
if (x3 > last) {
count++;
last = x3;
}
}

}
return count;
}

/**
* locate num in a sorted array
* @param num number to locate
* @param arr sorted array in asc order
* @return index of the num in array.If not find,-1 will be return.
*/
public static int locateNumInSortedArray(int num, int[] arr) {
if (arr.length == 0 || arr[0] > num || arr[arr.length - 1] < num) {
return -1;
}

int startIndex = 0, endIndex = arr.length - 1;
while (startIndex <= endIndex) {
int middleIndex = (startIndex + endIndex) >> 1;
if (num == arr[middleIndex]) {
return middleIndex;
} else if (num > arr[middleIndex]) {
startIndex = middleIndex + 1;
} else {
endIndex = middleIndex - 1;
}
}
return -1;
}
}
```

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