線性時間求最大間隔

 題目： 給定一個實數數組（無序），要求用線性時間求數軸上相鄰兩數間的最大間隔

關鍵：將除最大最小的n-2個數（或小於n-2） 投入n-1個桶中，這裏利用的鴿巢原理，至少一個桶是空的。

#include<iostream>
using namespace std;

void print(string, float *, int);

float data[] = {3.4, 2.1, 5.1, 1.4, 8.5, 7.4, 6.3};

float getMax(float data[], int size){
    int i;
    float max = data[0];
    for (i = 1; i < size; ++i) {
        if (data[i]>max)
            max = data[i];
    }
    return max;
}

float getMin(float data[], int size){
    int i;
    float min = data[0];
    for (i = 1; i < size; ++i) {
        if (data[i]<min)
            min = data[i];
    }
    return min;
}

float findMaxGap(float data[], int size){
    float max = getMax(data, size);
    float min = getMin(data, size);
    cout<<max<<" "<<min<<endl;
    int slot = size-1;
    float gap = (max-min)/slot;
    float *high = new float[slot];
    float *low = new float[slot];
    for (int i = 0; i < slot; ++i) {
        high[i] = min;
        low[i] = max;
    }

    for (int j = 0; j < size; ++j) {
        if(data[j]!=max&& data[j]!=min){
            int index = (data[j] - min)/gap;
            if(high[index]<data[j])
                high[index]= data[j];
            if(low[index]>data[j])
                low[index]=data[j];
        }
    }

    print("high array", high, slot);
    print("low  array", low,slot);

    float maxGap = 0;
    float lowedg = min;
    for (int var = 0; var < slot; ++var) {
        if(high[var]!=min){
            if(low[var]-lowedg>maxGap)
                maxGap = low[var]-lowedg;
            lowedg = high[var];
        }
    }

    return maxGap;
}