leetcode-Container With Most Water
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题目
Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container.
一开始的代码
nt maxArea(int* height, int heightSize) { int max = 0; if(heightSize == 0 || heightSize == 1){ return 0; } for(int i = 0; i < heightSize; i++){ for(int j = i + 1; j < heightSize; j++){ int row = j - i; int col = height[j] < height[i] ? height[j] : height[i]; int area = row * col; max = max > area ? max : area; } } return max;}复杂度为O(N2)超时了
2.0版本，还是超时，不过不是之前的超时用例了
int maxArea(int* height, int heightSize) { int max = 0; if(heightSize == 0 || heightSize == 1){ return 0; } for(int i = 0; i < heightSize; i++){ if(i <= heightSize / 2){//距离右端远 int j = heightSize - 1; while(j > i){//从右开始，找到高度大于i的第一个也是最远的j if(height[j] >= height[i]){ //printf("%d * %d | ", height[i], j -i); max = max > (height[i] * (j - i)) ? max : height[i] * (j - i); break; } else{ j--; } } if(j - i < i){//右端的距离小于左端最远的距离，则左端可能存在大于目前面积的列，检查左端 int k = 0; while(k < i && i - k > j - i){ if(height[k] >= height[i]){ //printf("%d * %d | ", height[i], i-k); max = max > (height[i] * (i - k)) ? max : height[i] * (i - k); break; } else{ k++; } } } } else{//距离左端远，同上 int j = 0; while(j < i){ if(height[j] >= height[i]){ //printf("%d * %d | ", height[i], i-j); max = max > (height[i] * (i - j)) ? max : height[i] * (i - j); break; } else{ j++; } } if(i - j < heightSize - i - 1){ int k = heightSize - 1; while(k > i && k - i > i - j){ if(height[k] >= height[i]){ //printf("%d * %d | ", height[i], k -i); max = max > (height[i] * (k - i)) ? max : height[i] * (k - i); break; } else{ k--; } } } } } return max;}3.0版，想到在上个版本上更进一步的简便方法，循环时多加个判断条件，可以少判断几次
int maxArea(int* height, int heightSize) { int max = 0; if(heightSize == 0 || heightSize == 1){ return 0; } for(int i = 0; i < heightSize; i++){ if(height[i] == 0){ continue; } if(i <= (heightSize - 1) / 2){//距离右端远 int j = heightSize - 1; while(j > i && ((j - i) > max / height[i])){//从右开始，找到高度大于i的第一个也是最远的j if(height[j] >= height[i]){ //printf("%d * %d | ", height[i], j -i); max = max > (height[i] * (j - i)) ? max : height[i] * (j - i); break; } else{ j--; } } if(j - i < i){//右端的距离小于左端最远的距离，则左端可能存在大于目前面积的列，检查左端 int k = 0; while(k < i && i - k > j - i){ if(height[k] >= height[i]){ //printf("%d * %d | ", height[i], i-k); max = max > (height[i] * (i - k)) ? max : height[i] * (i - k); break; } else{ k++; } } } } else{//距离左端远，同上 int j = 0; while(j < i && ((i - j) > max / height[i])){ if(height[j] >= height[i]){ //printf("%d * %d | ", height[i], i-j); max = max > (height[i] * (i - j)) ? max : height[i] * (i - j); break; } else{ j++; } } if(i - j < heightSize - i - 1){ int k = heightSize - 1; while(k > i && k - i > i - j){ if(height[k] >= height[i]){ //printf("%d * %d | ", height[i], k -i); max = max > (height[i] * (k - i)) ? max : height[i] * (k - i); break; } else{ k--; } } } } } return max;}还是未通过，当测试用例为倒序时，遍历的次数还是会较多，因此改下算法，先算下用例大概为升序还是降序，若为升序则从左向右开始遍历，若为降序，则从右开始向左遍历。
改完了，还是未过…只好根据测试用例，采用暴力破解了
int maxArea(int* height, int heightSize) { int max = 0; if(heightSize == 0 || heightSize == 1){ return 0; } int asc = 1; for(int i = 1; i < heightSize; i++){ if(height[i] < height[i - 1]){ asc = 0; break; } } if(asc == 1){//升序数列 for(int i = 0; i < heightSize; i++){ max = max > (heightSize - i - 1) * height[i] ? max : (heightSize - i - 1) * height[i]; } return max; } int desc = 1; for(int i = 0; i < heightSize; i++){ if(i == heightSize - 1){ if(height[i] > height[i - 1]){ desc = 0; break; } } else{ if(height[i] < height[i + 1]){ desc = 0; break; } } } if(desc == 1){ for(int i = heightSize - 1; i >= 0; i--){ max = max > i * height[i] ? max : i * height[i]; } return max; } if(asc >= 0){ for(int i = 0; i < heightSize; i++){ if(height[i] == 0){ continue; } if(i <= (heightSize - 1) / 2){//距离右端远 int j = heightSize - 1; while(j > i && ((j - i) > max / height[i])){//从右开始，找到高度大于i的第一个也是最远的j if(height[j] >= height[i]){ //printf("%d * %d | ", height[i], j -i); max = max > (height[i] * (j - i)) ? max : height[i] * (j - i); break; } else{ j--; } } if(j - i < i){//右端的距离小于左端最远的距离，则左端可能存在大于目前面积的列，检查左端 int k = 0; while(k < i && i - k > j - i){ if(height[k] >= height[i]){ //printf("%d * %d | ", height[i], i-k); max = max > (height[i] * (i - k)) ? max : height[i] * (i - k); break; } else{ k++; } } } } else{//距离左端远，同上 int j = 0; while(j < i && ((i - j) > max / height[i])){ if(height[j] >= height[i]){ //printf("%d * %d | ", height[i], i-j); max = max > (height[i] * (i - j)) ? max : height[i] * (i - j); break; } else{ j++; } } if(i - j < heightSize - i - 1){ int k = heightSize - 1; while(k > i && k - i > i - j){ if(height[k] >= height[i]){ //printf("%d * %d | ", height[i], k -i); max = max > (height[i] * (k - i)) ? max : height[i] * (k - i); break; } else{ k--; } } } } } } else{printf("new algo"); for(int i = heightSize - 1; i >= 0 ; i--){ if(height[i] == 0){ continue; } if(i >= (heightSize - 1) / 2){ int j = 0; while(j < i && (i - j) > max / height[i]){ if(height[j] >= height[i]){ max = max > (height[i] * (i - j)) ? max : height[i] * (i - j); break; } else{ j++; } } if(i - j < heightSize - i - 1){ int k = heightSize - 1; while(k > i && k - i > i - j){ if(height[k] >= height[i]){ max = max > (height[i] * (k - i)) ? max : height[i] * (k - i); break; } else{ k--; } } } } else{ int j = heightSize - 1; while(j > i && ((j - i) > max / height[i])){//从右开始，找到高度大于i的第一个也是最远的j if(height[j] >= height[i]){ //printf("%d * %d | ", height[i], j -i); max = max > (height[i] * (j - i)) ? max : height[i] * (j - i); break; } else{ j--; } } if(j - i < i){//右端的距离小于左端最远的距离，则左端可能存在大于目前面积的列，检查左端 int k = 0; while(k < i && i - k > j - i){ if(height[k] >= height[i]){ //printf("%d * %d | ", height[i], i-k); max = max > (height[i] * (i - k)) ? max : height[i] * (i - k); break; } else{ k++; } } } } } } return max;}观察测试用例，发现超时的用例要么是升序，要么是降序，因此，在最开始判断用例是升序还是降序，然后相应的遍历一次解决…
嗯，这是个投机取巧的方法，纯粹只是为了通过题目而已..
不过根据原来的算法，其实也才300ms左右，也算过了吧~哈哈
代码中我觉的有大量冗余的地方，但是懒得改了，就放在那里了
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