leetcode53：求解最大子序和（兩種方法）

算法思路和算法複雜度分析在函數中

public class leetcode_53 {
	/**
	 * 最大子序列和
	 * 
	 * 
	 */
	
	public static void main(String[] args) {
		//測試第一種思路：
		int[] array = {-2,-1};
		System.out.println(maxSubArray(array));
		
		//測試第二種思路：
		System.out.println(maxSubArray3(array));
	}
	
	
    public static int maxSubArray(int[] nums) {
    	
    	/**
    	 * 
    	 * 算法思路：maxNumber：全局最大值
    	 *        currentNumber : 當前最大值
    	 *          如果 currentNumber <=0 將currentNumber= nums[i];
    	 *     如果currentNumber> 0 將currentNumber += nums[i]
    	 */
    	
    	/**
    	 * 
    	 * 算法複雜度分析：O(n)
    	 * 
    	 * 
    	 */
    	int maxNumber = nums[0];
    	int currentNumber = nums[0];
    			
    	int len = nums.length;
    	for(int i =1;i<len;i++) {
    		if(currentNumber >0) {
    			currentNumber += nums[i];
    		}else {	
    			currentNumber = nums[i];
    		}
    		if(currentNumber > maxNumber) {
    			maxNumber = currentNumber;
    		}
    	}
		return maxNumber;
    }
	
    public static int maxSubArray3(int[] nums) {
    	return maxSubArray2(nums, 0, nums.length-1);
    }
    
    
    public static int maxSubArray2(int[] nums,int start ,int end) {
    	
    	/**
    	 * 算法思路：可以通過分治法進行求解，
    	 * leftmax：左邊一半最大值
    	 * rightmax ： 右邊一半最大值
    	 * leftsum+rightsum ： 最大值跨越middle
    	 * 
    	 * leftmax，rightmax通過遞歸
    	 * leftsum+rightsum通過O(N複雜度進行求解)
    	 * 
    	 */
    	
    	/**
    	 * 算法複雜度分析：遞歸表達式爲T(N) = 2T(N/2) + O(N)
    	 *                通過主定理，可以得出爲case2,所以算法複雜度爲O(NlgN)
    	 */

    	int leftmax,rightmax;
    	//int leftsum ,rightsum;
    	
    	if(start == end) {
    		return nums[start];
    	}else if(start>end){
    		return Integer.MIN_VALUE;
    	}
    	
    	int middle = (start+end)>>1;
    	leftmax = maxSubArray2(nums,start,middle); 
    	rightmax = maxSubArray2(nums,middle+1,end);
    	
    	int currentleftSum = 0;
    	int leftsum = Integer.MIN_VALUE;
    	for(int i = middle;i>=start;i--) {
    		currentleftSum  += nums[i];
    		if(currentleftSum > leftsum) {
    			leftsum = currentleftSum;
    		}
    	}
    	
    	int currentrightSum = 0;
    	int rightsum = Integer.MIN_VALUE;
    	for(int i = middle+1;i<=end;i++) {
    		currentrightSum += nums[i];
    		if(currentrightSum>rightsum) {
    			rightsum = currentrightSum;
    		}
    	}
    	
		return max3(leftmax,rightmax,leftsum+rightsum);
    }

    public static int max3(int x,int y,int z) {
    	int max = x>y?x:y;
    	return max>z?max:z;
    	
    }

}