LeetCode Trapping Rain Water

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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;    }};

先求出各個block的峰值，然後雨水肯定落在峰值block之間，對這些峰值block為界限的區間進行合并（對於序列A{4,1,3,1,5},第一個區間為A(0,2)，第二個區間為A(2,4)，由於A[4] > A[2] 且 A[0] > A[2]所以可以合并為區間A(0, 4)），直到不能繼續合并，最終計算這些block區間所能積累的水量。

再稍稍改進一下，合并過程在求峰值時同時進行，額外數組減少到一個

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;            }        }    }};

 

另外一種簡潔方法參見：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;    }};