一个有序数组（从小到大排列），数组中的数据有正有负，求这个数组中的最小绝对值

Find the min absolute value

  9 /*
 10  * binary search
 11  */
 12 int findMin_binary(int arr[], int left, int right){
 13     if (NULL == arr)
 14             return 0;
 15
 16     while(left < right){
 17             if (arr[left] > 0)
 18                     return arr[left];
 19             if (arr[right] < 0)
 20                     return arr[right];
 21
 22             if (right - left == 1){
 23                     return abs(arr[left]) > abs(arr[right])? (arr[right]) : (arr[left]);
 24             }
 25
 26             int mid = (left & right) + ((left ^ right) >> 1);
 27             if (arr[mid] > 0){
 28                     right = mid;
 29             }else{
 30                     left = mid;
 31             }
 32     }
 33 }
 35 /*
 36  * Ternary search  http://en.wikipedia.org/wiki/Ternary_search
 37  */
 38 int findMin_ternary(int arr[], int left, int right){
 39     if (NULL == arr)
 40             return 0;
 41
 42     while(left < right){
 43             if (arr[left] > 0)
 44                     return arr[left];
 45             if (arr[right] < 0)
 46                     return arr[right];
 47
 48             if (right - left == 2){
 49                     int mid = (left + right) >> 1;
 50                     int less_min = abs(arr[left]) > abs(arr[right])? (arr[right]) : (arr[left]);
 51                     return abs(arr[mid]) > abs(less_min) ? less_min : arr[mid];
 52             }
 53
 54             int left_third = (2*left + right)/3;
 55             int right_third = (left + 2*right)/3;
 56             if(abs(arr[left_third]) > abs(arr[right_third]))
 57                     left = left_third;
 58             else
 59                     right = right_third;
 60     }
 61 }

double phi = (1 + Math.sqrt(5)) / 2;
double resphi = 2 - phi;
 
// a and c are the current bounds; the minimum is between them.
// b is a center point
// f(x) is some mathematical function elsewhere defined
// a corresponds to x1; b corresponds to x2; c corresponds to x3
// x corresponds to x4
public double goldenSectionSearch(double a, double b, double c, double tau) {
    double x;
    if (c - b > b - a)
      x = b + resphi * (c - b);
    else
      x = b - resphi * (b - a);
    if (Math.abs(c - a) < tau * (Math.abs(b) + Math.abs(x))) 
      return (c + a) / 2; 
    assert(f(x) != f(b));
    if (f(x) < f(b)) {
      if (c - b > b - a) return goldenSectionSearch(b, x, c, tau);
      else return goldenSectionSearch(a, x, b, tau);
    }
    else {
      if (c - b > b - a) return goldenSectionSearch(a, b, x, tau);
      else return goldenSectionSearch(x, b, c, tau);
    }
  }