Trapping Rain Water
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【题目】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!
【题意】给定n个非负整数,代表一个柱状图,每一个柱子的宽度为1,计算下雨之后柱状图能装多少水?【解析】使用两个指针初始值是数组的起始位置和结束为值,用来维护装水两边的位置。当左边的高度低的时候,说明其右侧装的水和它一样高,此时左边的指针逐渐右移,并且同时加上左边指针与右移后的高度差,因为这里是可以装水的,当遇到同样高或者更高的位置时进行下一轮判断,同样右边指针也是同样的办法,最后知道左边指针和右边指针相遇停止。
实现:
public int trap(int[] A,int n) { int count=0,left=0,right=n-1; while(left<right) { int min=min(A[left],A[right]); if(A[left]==min) while(++left<right&&A[left]<min) count+=min-A[left]; else while(left<--right&&A[right]<min) count+=min-A[right]; } return count; }
时间复杂度:O(n)
空间复杂度:O(1)
别人的解法:首先找到有效的两边,如 [0, 1, 2, 3, 0, 3, 2, 1, 0],递增的左边和递减的右边是不能盛水的。
然后选择较短的边,如果是左边就开始右移,直到找到一个比左边高的边,更新左边;如果是右边较短,就左移,直到找到一个更高的边,更新右边。
public int trap(int[] A) { if (A.length < 3) return 0; int ans = 0; int l = 0, r = A.length - 1; // find the left and right edge which can hold water while (l < r && A[l] <= A[l + 1]) l++; while (l < r && A[r] <= A[r - 1]) r--; while (l < r) { int left = A[l]; int right = A[r]; if (left <= right) { // add volum until an edge larger than the left edge while (l < r && left >= A[++l]) { ans += left - A[l]; } } else { // add volum until an edge larger than the right volum while (l < r && A[--r] <= right) { ans += right - A[r]; } } } return ans; }
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