时间复杂度完美例子---数组最大子序列和

package 分而治之;

import utils.Utils;

public class 数组最大子序列和 {
    //time O(n^3) space O(1)
    public static int getMaxSumOfSequence01(int[] a){
        int maxsofar = 0;
        int startIndex = 0,endIndex = 0;
        for(int i=0;i<a.length;i++){
            for(int j=0;j<a.length;j++){
                int sum = 0;
                for(int k=i;k<=j;k++)
                    sum+=a[k];
                if(maxsofar<=sum){
                    startIndex = i;
                    endIndex = j;
                }
                maxsofar = Math.max(sum, maxsofar);
            }
        }
        System.out.println("["+startIndex+","+endIndex+"]");
        return maxsofar;
    }

    //time O(n^2) space O(1)
    public static int getMaxSumOfSequence02(int[] a){
        int maxsofar = 0;
        int startIndex = 0,endIndex = 0;
        for(int i=0;i<a.length;i++){
            int sum = 0;
            for(int j=i;j<a.length;j++){
                sum+=a[j];
                if(maxsofar<=sum){
                    startIndex = i;
                    endIndex = j;
                }
                maxsofar = Math.max(sum, maxsofar);
            }
        }
        System.out.println("["+startIndex+","+endIndex+"]");
        return maxsofar;
    }

    //time O(n) space O(1)
    public static int getMaxSumOfSequence04(int[] a){
        int maxsofar = 0;
        int sum = 0;
        int startIndex = 0,endIndex = 0;
        for(int i=0;i<a.length;i++){
            sum+=a[i];
            if(sum<=0) {
                sum = 0;
                startIndex = i+1;
            } else {
                if(maxsofar<=sum){
                    endIndex = i;
                }
                maxsofar = Math.max(sum, maxsofar);
            }
        }
        System.out.println("["+startIndex+","+endIndex+"]");
        return maxsofar;
    }

    //time O(n^2) space O(n)
    public static int getMaxSumOfSequence03(int[] a){
        int maxsofar = 0;
        int startIndex = 0,endIndex = 0;
        int[] currnum = new int[a.length];
        currnum[0] = a[0];
        for(int i=1;i<a.length;i++)
            currnum[i] = currnum[i-1]+a[i];
        for(int i=0;i<a.length;i++){
            int sum = 0;
            for(int j=i;j<a.length;j++){
                if(i==0) sum = currnum[j];
                else sum = currnum[j] - currnum[i-1];
                if(maxsofar<=sum){
                    startIndex = i;
                    endIndex = j;
                }
                maxsofar = Math.max(sum, maxsofar);
            }
        }
        System.out.println("["+startIndex+","+endIndex+"]");
        return maxsofar;
    }

    //O(log(n))
    public static int getMaxSumOfSequeneceByRecursion(int[] a,int l,int r)
    {
        int lmax,rmax,sum=0;
        int mid = (l+r)/2;
        if(l>r) //zero item
            return 0;
        if(l==r) //only one item
            return Math.max(a[l], 0);
        lmax = sum = 0;
        for(int i=mid;i>=l;i--){
            sum+=a[i];
            lmax = Math.max(lmax, sum);
        }
        rmax = sum = 0;
        for(int i=mid+1;i<=r;i++){
            sum+=a[i];
            rmax = Math.max(rmax, sum);
        }
        return Utils.maxOfThree(lmax+rmax,getMaxSumOfSequeneceByRecursion(a,l,mid),getMaxSumOfSequeneceByRecursion(a,mid+1,r));
    }

    public static void main(String[] args) {
        //int a[] = {31,-41,59,26,-53,58,97,-93,-23,84};
        int a[] = Utils.produceNum(-10000, 10000, 10000);
        long t1 = System.nanoTime();
        System.out.println(getMaxSumOfSequence01(a));
        long t2 = System.nanoTime();
        System.out.println("m1 time:"+(t2-t1)/1000);
        System.out.println(getMaxSumOfSequence02(a));
        long t3 = System.nanoTime();
        System.out.println("m2 time:"+(t3-t2)/1000);
        System.out.println(getMaxSumOfSequence03(a));
        long t4 = System.nanoTime();
        System.out.println("m3 time:"+(t4-t3)/1000);
        System.out.println(getMaxSumOfSequence04(a));
        long t5 = System.nanoTime();
        System.out.println("m4 time:"+(t5-t4)/1000);
        System.out.println(getMaxSumOfSequeneceByRecursion(a,0,a.length-1));
        long t6 = System.nanoTime();
        System.out.println("m5递归 time:"+(t6-t5)/1000);
    }
}