[leetcode#240] Search a 2D Matrix II

problem:

Write an efficient algorithm, searches for a value in a m x n Matrix. This matrix has the following properties:

• Integers in each row is sorted in ascending from left to right.
• Integers in each column is sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[  [1,   4,  7, one, a],  [2,   5, 8,, +  ],  [3,   6,  9, +, +],  [10, 13, 14, 17,  [18, 21, 23, 26, 30]

Given target = 5 , return true .

Given target = 20 , return false .

Analysis 1:

This problem was still very easy, but it reminds me to repick up some skills I had abandoned forALongTime . Instant idea:since Each column and row are sorted, we can take advantage of one of the them. Go through each row one by one, forEach row, we perform binary search. Note:pitfall in binary search.1. Wrong stop condition. while(Left <Right ) input:[[-5]],-5Output:falseexpected:trueStart:left= 0, right = 0Not able to enter the loop, not able to reachintMid = (left + right)/2;if(Nums[mid] = =target)return true; Fix: while(Left <=Right )2. Wrong policy in the update left and right pointer.if(Nums[mid] = =target)return true;Else if(Nums[mid] >target) { Right=Mid;} Else{ Left=Mid;} input:[[-1,3]], 1Start:left= 0, right = 1, enter Loop:mid= 0; Nums[mid]<Target Left= Mid = 0. No updates at all, would leads to infinite loop. Binary search is very important, you must remember the coding structure clearly. Time Complexity:o (NLOGN)

Solution 1:

 Public classSolution { Public BooleanSearchmatrix (int[] Matrix,inttarget) {        if(Matrix = =NULL)            Throw NewIllegalArgumentException ("The reference of the matrix is null"); if(Matrix.length = = 0 | | matrix[0].length = = 0)            return false;  for(inti = 0; i < matrix.length; i++) {            if(BinarySearch (matrix[i], target))return true; }        return false; }            Private BooleanBinarySearch (int[] Nums,inttarget) {        if(Nums[0] > Target | | target > NUMS[NUMS.LENGTH-1])            return false; intleft = 0; intright = Nums.length-1;  while(Left <=Right ) {            intMid = (left + right)/2; if(Nums[mid] = =target)return true; Else if(Nums[mid] >target) { Right= Mid-1; } Else{ Left= Mid + 1; }        }        return false; }}

Analysis 2:

Even though we have take advantage of the sorted property in each row, what about each column?Can We also take advantage of it?So as to reduce time complexity. Sure!!!great Idea:start from up-right-Most corner, then use the sorted properties to exclude elements. Principle:1. IFF Matrix[x][y] >target, since Y is sorted, target must isn't able to appear on y column, we should skip it. while(Y >= 1 && Target <Matrix[x][y]) {y--;} 2. IFF Matrix[x][y] <target.since matrix[x][y-1] must smaller than it, and matrix[x][y+1] is impossilbe. (We have do in the previous step) the only possible direction is below. use ThisInvariant,ifthe target exist in the matrix, we must is able to find it. for(intx = 0; x < matrix.length; X + +) {     while(Y >= 1 && Target <Matrix[x][y]) {y--; }       if(Matrix[x][y] = =target)return true;}

soltuion 2:

 Public classSolution { Public BooleanSearchmatrix (int[] Matrix,inttarget) {        if(Matrix = =NULL)            Throw NewIllegalArgumentException ("The reference of the matrix is null"); if(Matrix.length = = 0 | | matrix[0].length = = 0)            return false; inty = matrix[0].length-1;  for(intx = 0; x < matrix.length; X + +) {             while(Y >= 1 && Target <Matrix[x][y]) {y--; }               if(Matrix[x][y] = =target)return true; }        return false; }}

[leetcode#240] Search a 2D Matrix II