首页 > 代码库 > LeetCode Trapping Rain Water
LeetCode Trapping Rain Water
class Solution {public: int trap(int A[], int n) { if (n < 1) return 0; vector<int> peaks; vector<int> peaks_tmp; int last_idx = 0; bool increasing = true; for (int i=1; i<n; i++) { if (A[i] < A[last_idx]) { if (increasing) { peaks.push_back(last_idx); } increasing = false; } else { increasing = true; } last_idx = i; } if (increasing) peaks.push_back(n - 1); if (peaks.size() < 2) return 0; bool updated = true; while (updated) { updated = false; peaks_tmp.clear(); peaks_tmp.push_back(peaks[0]); peaks_tmp.push_back(peaks[1]); for (int i=2; i<peaks.size(); i++) { int tlen = peaks_tmp.size(); int ai = peaks_tmp[tlen - 2]; int bi = peaks_tmp[tlen - 1]; int ci = peaks[i]; if (A[ai] >= A[bi] && A[ci] >= A[bi]) { peaks_tmp[tlen - 1] = ci; updated = true; } else { peaks_tmp.push_back(ci); } } swap(peaks, peaks_tmp); } int rain = 0; for (int i=1; i<peaks.size(); i++) { int left = peaks[i - 1]; int right= peaks[i]; int h = min(A[left], A[right]); int blocks = 0; for (int i=left + 1; i<right; i++) blocks += A[i] > h ? h : A[i]; rain += h * (right - left - 1) - blocks; } return rain; }};
先求出各个block的峰值，然后雨水肯定落在峰值block之间，对这些峰值block为界限的区间进行合并（对于序列A{4,1,3,1,5},第一个区间为A(0,2)，第二个区间为A(2,4)，由于A[4] > A[2] 且 A[0] > A[2]所以可以合并为区间A(0, 4)），直到不能继续合并，最终计算这些block区间所能积累的水量。
再稍稍改进一下，合并过程在求峰值时同时进行，额外数组减少到一个
class Solution {public: int trap(int A[], int n) { if (n < 1) return 0; vector<int> peaks; int last_idx = 0; bool increasing = true; for (int i=1; i<n; i++) { if (A[i] < A[last_idx]) { if (increasing) { peaks.push_back(last_idx); compact(A, peaks); } increasing = false; } else { increasing = true; } last_idx = i; } if (increasing) peaks.push_back(n - 1); compact(A, peaks); if (peaks.size() < 2) return 0; int rain = 0; for (int i=1; i<peaks.size(); i++) { int left = peaks[i - 1]; int right= peaks[i]; int h = min(A[left], A[right]); int blocks = 0; for (int i=left + 1; i<right; i++) blocks += A[i] > h ? h : A[i]; rain += h * (right - left - 1) - blocks; } return rain; } void compact(int A[], vector<int> &peaks) { int len = peaks.size(); if (len < 3) return; bool updated = true; while (updated && (len = peaks.size()) >= 3) { updated = false; int ai = peaks[len - 3]; int bi = peaks[len - 2]; int ci = peaks[len - 1]; if (A[ai] >= A[bi] && A[ci] >= A[bi] ) { peaks[len - 2] = ci; peaks.pop_back(); updated = true; } } }};
另外一种简洁方法参见：http://www.cnblogs.com/zhuli19901106/p/3542216.html，如下
class Solution {public: int trap(int A[], int n) { if (n < 2) return 0; vector<int> lmax(n, 0); vector<int> rmax(n, 0); int m = A[n-1]; for (int i=n-2; i >= 0; i--) { if (A[i] > m) m = A[i]; rmax[i] = m; } m = A[0]; for (int i=1; i<n; i++) { if (A[i] > m) m = A[i]; lmax[i] = m; } int rain = 0; for (int i=0; i<n; i++) { int h = min(lmax[i], rmax[i]); int v = h - A[i]; rain += v < 0 ? 0 : v; } return rain; }};
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。