LeetCode Trapping Rain Water

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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!

思路：如果我们能做到每次到每个位置的时候就能得到答案的话，那么复杂度就低了，所以我们可以想如果每次存在一个高的紧接着是一个相对矮的话，就能计算了，但是我们还要确保它的右边也能装下，所以我们先找到最高的那个，然后分成左右两边计算，再多一个curMax就能按照前面的思路计算了

class Solution {
public:
    int trap(int A[], int n) {
		int ans = 0;
		int maxIndex = 0;
		for (int i = 1; i < n; i++) 
			if (A[i] > A[maxIndex])
				maxIndex = i;

		int curMax = A[0];
		for (int i = 1; i < maxIndex; i++) {
			if (A[i] > curMax)
				curMax = A[i];
			else ans += curMax - A[i];
		}
		curMax = A[n-1];
		for (int i = n-2; i > maxIndex; i--) {
			if (A[i] > curMax)
				curMax = A[i];
			else ans += curMax - A[i];
		}

		return ans;
    }
};

LeetCode Trapping Rain Water

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原文地址：http://blog.csdn.net/u011345136/article/details/44081033