[leetcode] Trapping Rain Water
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思路1:对某个值A[i]来说,能装的最多的水取决于在i之前最高的值leftMax[i]和在i右边的最高的rightMax[i],容水量即min(leftMax[i],rightMax[i]) – A[i]。:为了计算高度,第一遍从左到右计算数组leftMaxt,第二遍从右到左计算rightMax。时空复杂度都是O(n)。
int trap(int A[], int n)
{
if(n == 0 || n == 1 || n==2)
return 0;
int *leftMax = new int[n];
int *rightMax = new int[n];
int area = 0;
int max;
max = A[0];
for(int i = 0; i < n; i++)
{
leftMax[i] = max;
if(A[i] > max)
max = A[i];
}
max = A[n-1];
for(int i = n-1; i >= 0; i--)
{
rightMax[i] = max;
if(A[i] > max)
max = A[i];
}
for(int i = 0; i < n; i++)
{
int h = min(leftMax[i], rightMax[i]);
if(A[i] < h)
area += h-A[i];
}
return area;
}
思路2:改进思路1,可以用常量空间搞定。具体思路,先找到最高的柱,设下标为maxIdx,然后由两边向中间计算,边计算边更新当前的最高高度。
int trap(int A[], int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int water = 0;
int maxIdx = 0;
for (int i = 1; i < n; i++) {
if (A[i] > A[maxIdx]) {
maxIdx = i;
}
}
int max = A[0];
for (int i = 1; i < maxIdx; i++) {
if (A[i] > max) {
max = A[i];
}
else {
water += max - A[i];
}
}
max = A[n - 1];
for (int i = n - 2; i > maxIdx; i--) {
if (A[i] > max) {
max = A[i];
}
else {
water += max - A[i];
}
}
return water;
}
// Start typing your C/C++ solution below
// DO NOT write int main() function
int water = 0;
int maxIdx = 0;
for (int i = 1; i < n; i++) {
if (A[i] > A[maxIdx]) {
maxIdx = i;
}
}
int max = A[0];
for (int i = 1; i < maxIdx; i++) {
if (A[i] > max) {
max = A[i];
}
else {
water += max - A[i];
}
}
max = A[n - 1];
for (int i = n - 2; i > maxIdx; i--) {
if (A[i] > max) {
max = A[i];
}
else {
water += max - A[i];
}
}
return water;
}
思路3:设置两个指针,一个从最左边开始遍历,一个从最右边开始遍历。只要两个指针不相遇,则从较低侧往中间遍历,当
堤坝高度小于或等于这个较低侧高度时,将高度差加到蓄水和。
int trap2(int A[], int n)
{
if(n == 0 || n == 1 || n==2)
return 0;
int left = 0;
int right = n-1;
int area = 0;
int leftH = 0;
int rightH = 0;
while(left < right)
{
if(A[left] < A[right])
{
if(A[left] < leftH)
area += leftH - A[left];
else
leftH = A[left];
left++;
}
else
{
if(A[right] < rightH)
area += rightH - A[right];
else
rightH = A[right];
right--;
}
}
return area;
}
{
if(n == 0 || n == 1 || n==2)
return 0;
int left = 0;
int right = n-1;
int area = 0;
int leftH = 0;
int rightH = 0;
while(left < right)
{
if(A[left] < A[right])
{
if(A[left] < leftH)
area += leftH - A[left];
else
leftH = A[left];
left++;
}
else
{
if(A[right] < rightH)
area += rightH - A[right];
else
rightH = A[right];
right--;
}
}
return area;
}
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