135. Candy

First, the topic

1, examining

　　

2. Analysis

Each child corresponds to a grade, each child at least one sugar, next to the higher level of the child than next to the lower level of the candy more, beg at least to divide a few sweets.

Second, the answer

1, Ideas:

Method One,

Consider the number of sweets each child has to be affected by the child's rank on both sides of the left and right. So

①, find out all the children of the lowest level of the child 1, a sugar, and his left if there is a child 2 and the level is higher than the child 1, the child 2 candy for children 1 Sweets + 1, the final child 2 points of candy to be greater than equal to this number.

Similarly, the child 1 to the right of the companion child 3 if the level is greater than the child 1, also first assigned (Child 1) +1, the child 3 the final distribution of candy to be greater than the number of times.

②, the child 1 left all children follow the ① continue to recursion, the child 1 right child according to ① recursion.

③, and finally add all the sweets assigned to the child.

 Public intCandyint[] ratings) {                intLen =ratings.length; int[] Cost =New int[Len]; Candyhelper (ratings, cost,0, Len-1); intresult = 0;  for(inti = 0; i < Len; i++) Result+=Cost[i]; returnresult; }        Private voidCandyhelper (int[] Ratings,int[] Cost,intStartintend) {                if(Start >end)return; intMinindex = start;//rating the smallest subscript         for(intj = start + 1; J <= End; J + +) {            if(Ratings[minindex] >Ratings[j]) Minindex=J; } Cost[minindex]= Math.max (Cost[minindex], 1); if(minIndex-1 >= start && ratings[minindex-1]! = Ratings[minindex])//not equal to the leftCost[minindex-1] = Math.max (Cost[minindex-1], Cost[minindex] + 1); if(Minindex + 1 <= End && Ratings[minindex + 1]! = Ratings[minindex])//not equal to rightCost[minindex + 1] = Math.max (Cost[minindex + 1], Cost[minindex] + 1); Candyhelper (ratings, cost, start, Minindex+ W); Candyhelper (ratings, cost, Minindex+ 1, end); }

Method Two,

①, create a new cost array, the initial value is 1;

②, from the previous to traverse the ratings, if the latter element of the rating is greater than the previous one, then the cost of the last element of the previous value + 1

③, from the back forward to traverse the ratings, if the previous element's rating is greater than the latter, then the cost of the previous element is the maximum value of the current cost and the latter value + 1.

     Public intCandy2 (int[] ratings) {                intLen =ratings.length; if(Len < 2)            returnLen; int[] Cost =New int[Len]; Arrays.fill (Cost,1); //sure right higher rated child gets 1 more candy than left lower rated child         for(inti = 1; i < Len; i++) {            if(Ratings[i] > Ratings[i-1]) Cost[i]= Cost[i-1] + 1; }        //Scan from sure left higher rated child gets 1 more candy than right lower rated child         for(inti = len-1; i > 0; i--) {            if(Ratings[i-1] >ratings[i]) cost[i-1] = Math.max (Cost[i-1], cost[i] + 1); }                intresult = 0;  for(inti = 0; i < Len; i++) Result+=Cost[i]; returnresult; }

135. Candy