[LeetCode 42] Trapping Rain Water

标签：java   leetcode   算法   

题目链接：trapping-rain-water

import java.util.Stack;



/**
 * 
		Given n non-negative integers representing an elevation map where the width of each bar is 1, 
		compute how much water it is able to trap after raining.
		
		For example, 
		Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.
		
		
		The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. 
		In this case, 6 units of rain water (blue section) are being trapped. 
		Thanks Marcos for contributing this image!
 *
 */

public class TrappingRainWater {


      //解法一
//    315 / 315 test cases passed.
//    Status: Accepted
//    Runtime: 248 ms
//    Submitted: 0 minutes ago
    
    //时间复杂度O(n), 空间复杂度O(1)
    public int trap(int[] A) {
     	int sum = 0;	//
     	int max = 0;
     	
     	for (int i = 0; i < A.length; i++) { //找到最大值
			if(A[i] > A[max]) {
				max = i;
			}
		}
     	//计算左边
     	int topest = 0;
        for (int i = 0; i < max; i++) {
			if(A[i] > topest) {
				topest = A[i];
			} else {
				sum += topest - A[i];
			}
		} 
        //计算右边
        topest = 0;
        for (int i = A.length - 1; i > max; i--) {
			if(A[i] > topest) {
				topest = A[i];
			} else {
				sum += topest - A[i];
			}
		}
     	return sum;   	
     }
    
    //解法二
//    315 / 315 test cases passed.
//    Status: Accepted
//    Runtime: 282 ms
//    Submitted: 1 minute ago
    
    //时间复杂度O(n) 空间复杂度O(n)
    public int trap1(int[] A) {
		Stack<Integer> stack = new Stack<Integer>();
		int sum = 0;
		int i = 0;
		int max = 0;							//栈中的最大值
		for (; i < A.length - 1; i++) {			//找到最左边的最大值
			if (A[i] > A[i + 1]) {
				stack.add(A[i]);
				max = A[i];
				break;
			}
		}
		for (i = i + 1; i < A.length; i++) {
			int count = 1;
			while (!stack.isEmpty() && stack.peek() <= A[i]) {	//如果栈顶元素小于当前元素
				
				if(A[i] >= max) {//如果当前元素大于或等于栈中的最大值
					while(!stack.isEmpty()) {			//将栈中的元素全部出栈
						sum += max - stack.pop();
					}
					max = A[i];							//设置当前栈的最大值
					
				} else {		//如果当前元素小于栈中的最大值
					sum += A[i] - stack.pop();			
					count++;							//累计A[i]要进栈的次数
				}
			}
			
			while(count != 0) {
				stack.add(A[i]);
				count --;
			}
		}
		return sum;
	}
	public static void main(String[] args) {


	}

}

[LeetCode 42] Trapping Rain Water

标签：java   leetcode   算法   

原文地址：http://blog.csdn.net/ever223/article/details/44681121